Question 1

Which element in the modern periodic table has the electronic configuration 2, 8, 3 in its neutral atom?

Accepted Answer

Introduction / Context: Electronic configuration describes how electrons are distributed among the shells or energy levels of an atom. For lighter elements, school level chemistry often uses a simple notation such as 2, 8, 3 to represent electrons in the first, second and third shells. Recognising which element corresponds to a given configuration is important for understanding periodic trends, valency and chemical behaviour. This question asks you to identify the element whose neutral atom has the configuration 2, 8, 3. Given Data / Assumptions: The configuration 2, 8, 3 means 2 electrons in the first shell, 8 in the second and 3 in the third shell. The total number of electrons in a neutral atom equals its atomic number. The options include magnesium, sodium, aluminium, argon and neon. We assume the simple shell model often used up to calcium in school chemistry. Concept / Approach: To identify the element, we add the electrons in all shells in the configuration. The sum gives the atomic number, which can then be matched to the correct element name. The configuration 2, 8, 3 sums to 13, so we look for the element with atomic number 13, which is aluminium. Magnesium has atomic number 12, sodium has 11, neon has 10 and argon has 18. This straightforward calculation connects shell configuration with the periodic table entry. Step-by-Step Solution: Step 1: Add the electrons indicated in the configuration 2, 8, 3: 2 plus 8 plus 3 equals 13. Step 2: The total number of electrons in a neutral atom is equal to its atomic number. Step 3: Therefore, the atomic number corresponding to this configuration is 13. Step 4: Recall that the element with atomic number 13 in the modern periodic table is aluminium (symbol Al). Step 5: Check the other options: sodium has atomic number 11 with configuration 2, 8, 1; magnesium has 12 with configuration 2, 8, 2; argon has 18 with configuration 2, 8, 8; neon has 10 with configuration 2, 8. Step 6: None of the other elements match a total of 13 electrons, confirming aluminium as the correct choice. Verification / Alternative check: Using the periodic table, you can count positions from hydrogen upwards to confirm that aluminium is element 13. Aluminium is located in period 3 and group 13, which is consistent with an outer shell containing three electrons. Its valency is often three in compounds such as aluminium oxide and aluminium chloride, matching the three valence electrons implied by the configuration 2, 8, 3. These independent checks from periodic position and common valencies confirm that aluminium is the element with the given configuration. Why Other Options Are Wrong: Option A, magnesium, has atomic number 12 and configuration 2, 8, 2, with only two valence electrons. Option B, sodium, has atomic number 11 and configuration 2, 8, 1, with one valence electron. Option D, argon, is a noble gas with atomic number 18 and configuration 2, 8, 8, with a completely filled outer shell. Option E, neon, has atomic number 10 and configuration 2, 8, also with a filled outer shell. None of these match the specified configuration 2, 8, 3. Only option C, aluminium with atomic number 13, fits correctly. Common Pitfalls: Students sometimes make arithmetic mistakes when adding the electrons in each shell or they may misremember atomic numbers. Another common error is to think of magnesium when they see a configuration ending in 2 or to match argon incorrectly because it is also in period 3. To avoid these problems, always first sum the electron counts to get the atomic number, and then cross check this number with a periodic table or your memorised list of elements up to at least 20. This systematic approach reduces confusion and ensures accuracy. Final Answer: The element with electronic configuration 2, 8, 3 is Aluminium (atomic number 13), which has three valence electrons in its outermost shell.

Question 2

Approximately how many metalloids are commonly recognised in the modern periodic table in standard school level chemistry?

Accepted Answer

Introduction / Context: Metalloids form a small but important group of elements in the periodic table that show properties intermediate between metals and non metals. They are often used in semiconductors, glass, alloys and other specialised applications. At the school level, a standard list of metalloids is usually taught, and students are sometimes asked to recall how many such elements there are. This question focuses on the approximate count of commonly recognised metalloids in the modern periodic table as presented in typical textbooks. Given Data / Assumptions: Metalloids lie along the diagonal zigzag line separating metals and non metals in the periodic table. Commonly listed metalloids include boron, silicon, germanium, arsenic, antimony and tellurium. Some sources also include polonium or astatine, but school level lists often focus on the core group. We are concerned with the number used in standard general science and school chemistry materials. Concept / Approach: The most commonly taught list of metalloids at school level includes six elements: boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb) and tellurium (Te). These elements lie along the border between metals and non metals and exhibit intermediate properties. While some advanced sources debate adding more elements, the typical educational answer for how many metalloids there are is six. The approach is to recall this standard list and count its members, then select the option that matches this number. Step-by-Step Solution: Step 1: List the commonly accepted metalloids: boron, silicon, germanium, arsenic, antimony and tellurium. Step 2: Count these elements: boron is one, silicon two, germanium three, arsenic four, antimony five and tellurium six. Step 3: Some sources may also mention polonium or astatine, but in most school textbooks these are not included in the basic metalloid list. Step 4: Therefore, at this level, the number of metalloids is usually given as six. Step 5: Compare this count with the options and choose six metalloids as the correct answer. Verification / Alternative check: If you look at a periodic table with metal, metalloid and non metal colours indicated, you typically see a zigzag line with six elements highlighted as metalloids. Chemistry textbooks for secondary school also usually include a table where boron, silicon, germanium, arsenic, antimony and tellurium are listed together under the heading metalloids. They also illustrate their intermediate properties such as semiconducting behaviour and moderate electrical conductivity. These consistent references support the answer of six metalloids for exam purposes. Why Other Options Are Wrong: Option A, 5 metalloids, undercounts the usual group and would exclude one of the standard elements. Option B, 4 metalloids, is too low and does not reflect the usual educational list. Option C, 7 metalloids, might be used in some advanced contexts if elements like polonium or astatine are included, but this is beyond normal school level practice. Option E, 10 metalloids, is far above any commonly accepted count. Only option D correctly matches the widely taught number of six metalloids in the modern periodic table at this level. Common Pitfalls: Students may confuse the exact count because some periodic tables and references vary slightly, especially when they include borderline elements. Another pitfall is to misinterpret elements along the zigzag line and accidentally count non standard ones. To avoid confusion in school examinations, it is best to memorise the commonly listed six metalloids and their symbols. Remembering that they include boron and silicon, which are important in glass and electronics, along with germanium, arsenic, antimony and tellurium, will help you recall both their names and that there are six of them. Final Answer: The number of metalloids commonly recognised in the modern periodic table at school level is 6 metalloids.

Question 3

Which is the only non metal that is placed on the left side of the modern periodic table along with the alkali metals?

Accepted Answer

Introduction / Context: The modern periodic table is arranged so that metals, non metals and metalloids occupy characteristic regions. Most non metals are found on the right side of the table, while metals occupy the left and centre. However, there is one important exception among the non metals that is placed on the far left in group 1, above the alkali metals. This special positioning reflects both its electron configuration and its unique properties. The question asks you to identify this non metal that appears on the left side of the periodic table. Given Data / Assumptions: Group 1 of the periodic table is the alkali metal group, including lithium, sodium and potassium. Hydrogen is often shown above lithium in group 1, but it is not a metal under normal conditions. Most other non metals such as carbon, nitrogen, oxygen, halogens and noble gases are located on the right side of the table. The question is about position in the standard table layout, not specialised versions. Concept / Approach: Hydrogen has atomic number 1 and an electronic configuration of 1 electron in its first shell. This resembles the outer electron configuration of alkali metals, which each have one electron in their outermost shell. For this reason, hydrogen is placed in group 1 at the top. However, hydrogen is a non metal gas at room temperature and has many properties quite different from alkali metals. Carbon, neon, helium and chlorine are all non metals but are located on the right side of the table, not in group 1. The approach is to recognise that hydrogen is the only non metal placed in the leftmost column. Step-by-Step Solution: Step 1: Recall that the alkali metals (group 1) are lithium, sodium, potassium, rubidium, cesium and francium. Step 2: Above lithium in this group, the element usually shown is hydrogen with atomic number 1. Step 3: Hydrogen is a non metal gas under standard conditions and forms covalent molecules such as H2. Step 4: Carbon is located in group 14 near the middle right of the table, not in the leftmost column. Step 5: Neon and helium are noble gases in group 18 on the far right side of the periodic table. Step 6: Chlorine is a halogen in group 17, also on the right side. Step 7: Therefore, hydrogen is the only non metal positioned on the left side with the alkali metals. Verification / Alternative check: Looking at any standard periodic table chart confirms this layout. Group 1 shows hydrogen at the top followed by the alkali metals. Textbooks often describe hydrogen as a special case that does not fit neatly into any one group, but for convenience it is placed above the alkali metals. The non metals carbon, nitrogen, oxygen, halogens and noble gases form a cluster on the right side, separate from the main metallic block. These consistent representations across many sources confirm that hydrogen is the only non metal found on the leftmost side. Why Other Options Are Wrong: Option B, carbon, is a non metal but is located in the p block on the right side, not in the leftmost group. Option C, neon, and option D, helium, are noble gases in group 18, at the right edge of the table. Option E, chlorine, is a halogen in group 17 next to the noble gases. None of these occupy the leftmost column with the alkali metals. Only option A, hydrogen, is both a non metal and placed in group 1 on the left side of the periodic table. Common Pitfalls: Some students mistakenly assume that hydrogen must be a metal because it appears with metals in group 1. Others forget that group placement is based on electronic configuration and that hydrogen is unique in having properties similar to both alkali metals and halogens. To avoid confusion, remember that hydrogen is non metallic but is placed in group 1 mainly due to having one valence electron. This special status makes it an exception among non metals and explains its unusual position. Final Answer: The only non metal placed on the left side of the modern periodic table with the alkali metals is Hydrogen.

Question 4

Calculate the relative molecular mass of water (H2O) using standard atomic masses.

Accepted Answer

Introduction / Context: Relative molecular mass is an important concept in chemistry that links the atomic masses of elements to the mass of a molecule. It is used in calculations involving moles, gas laws and solution chemistry. Water is one of the most fundamental compounds, and students are often expected to know or quickly calculate its relative molecular mass. This question asks you to compute the relative molecular mass of H2O given standard atomic masses for hydrogen and oxygen. Given Data / Assumptions: The formula of water is H2O. Each molecule of water contains two hydrogen atoms and one oxygen atom. The approximate relative atomic mass of hydrogen is 1 u. The approximate relative atomic mass of oxygen is 16 u. We use simple whole number values for school level calculations. Concept / Approach: Relative molecular mass is found by adding the relative atomic masses of all atoms present in one molecule, taking into account how many of each atom there are. For water, the formula H2O indicates two hydrogens and one oxygen. We multiply the atomic mass of hydrogen by two, add the atomic mass of oxygen once and obtain the total. This simple procedure generalises to any molecular or formula mass calculation and is routinely used in stoichiometry. Step-by-Step Solution: Step 1: Identify the elements present in water H2O: hydrogen and oxygen. Step 2: Note that there are two hydrogen atoms and one oxygen atom in each water molecule. Step 3: Use the approximate relative atomic masses: hydrogen is 1 u, oxygen is 16 u. Step 4: Compute the contribution from hydrogen: 2 multiplied by 1 u equals 2 u. Step 5: Compute the contribution from oxygen: 1 multiplied by 16 u equals 16 u. Step 6: Add these contributions: 2 u plus 16 u equals 18 u. Step 7: Therefore, the relative molecular mass of water H2O is 18 u. Verification / Alternative check: If you use more precise atomic masses, such as 1.008 u for hydrogen and 16.00 u for oxygen, the calculation is slightly refined: 2 times 1.008 equals 2.016, plus 16.00 equals approximately 18.016 u. For most school level work, this is rounded to 18 u. This value is consistent with molar mass calculations where one mole of water has a mass of about 18 grams. Many chemistry tables and problem solutions explicitly state that H2O has a relative molecular mass of 18 u, confirming this result. Why Other Options Are Wrong: Option B, 1.8 u, is off by a factor of ten and might arise from misplacing a decimal point, not from correct atomic mass addition. Option C, 2.0 u, corresponds only to the hydrogen atoms and ignores the much heavier oxygen atom. Option D, 20 u, is too high and does not match any reasonable atomic mass combination for H2O. Option E, 28 u, might be confused with nitrogen gas N2, which has a relative molecular mass close to 28 u. None of these values result from correctly adding 2 times 1 u and 16 u. Only option A matches the correct calculation for water. Common Pitfalls: Students sometimes forget to multiply the atomic mass of hydrogen by two because there are two hydrogen atoms in the formula. Others may use incorrect atomic masses from memory or confuse water with other molecules. A reliable strategy is always to write out the formula, list each type of atom with its subscript, multiply each atomic mass by the number of atoms and then add the results. Practising this method on common compounds like water, carbon dioxide and methane builds confidence for more complex calculations later. Final Answer: The relative molecular mass of water H2O is 18 u, obtained by adding the masses of two hydrogens and one oxygen atom.

Question 5

A solution contains 33 g of common salt dissolved in 320 g of water; calculate the concentration of the solution in terms of mass by mass percentage.

Accepted Answer

Introduction / Context: This chemistry question tests the concept of expressing the concentration of a solution in terms of mass by mass percentage. Such questions are very common in school examinations and competitive tests because mass percentage is one of the most widely used ways to describe how much solute is present in a given amount of solution. Understanding this idea helps students in topics like stoichiometry, preparation of laboratory solutions, and even day to day applications such as reading labels on packaged food or medicines. Given Data / Assumptions: The solute is common salt, which we can treat as sodium chloride. Mass of solute (common salt) = 33 g. Mass of solvent (water) = 320 g. Total mass of the solution is equal to the sum of the masses of solute and solvent. The temperature and volume effects are ignored because we are working only with mass by mass percentage. Concept / Approach: Mass by mass percentage tells us how many grams of solute are present in 100 g of the final solution. The standard formula used is: mass percent of solute = (mass of solute / mass of solution) * 100 We first find the total mass of the solution by adding solute and solvent masses, and then substitute the values into the formula. All masses must be in the same units, which they already are here (grams). Step-by-Step Solution: Step 1: Calculate mass of solution. mass of solution = mass of solute + mass of solvent mass of solution = 33 g + 320 g = 353 g Step 2: Use the mass percent formula. mass percent of salt = (mass of solute / mass of solution) * 100 mass percent of salt = (33 / 353) * 100 Step 3: Perform the calculation approximately. 33 / 353 is about 0.09348 mass percent of salt ≈ 0.09348 * 100 ≈ 9.35% So the concentration of the solution in terms of mass by mass percentage is approximately 9.35 percent. Verification / Alternative Check: We can check if this value is sensible by estimating. The solute mass 33 g is a little less than one tenth of the total mass 353 g. One tenth of 353 g is 35.3 g, which would correspond to 10 percent. Since 33 g is slightly smaller than 35.3 g, the percentage should be slightly less than 10 percent. The calculated value 9.35 percent fits this rough estimation well, so the answer is consistent and reasonable. Why Other Options Are Wrong: Option B: 9.09% is based on an incorrect total mass or rounded fraction; it does not match the correct ratio 33 out of 353. Option C: 10.30% would imply more solute per 100 g of solution than we actually have, so it overestimates the concentration. Option D: 13.05% is far too large for 33 g of salt in 353 g of solution and would require more solute than is given. Common Pitfalls: Students often forget to include the solvent when calculating mass of solution and mistakenly divide by 320 g instead of 353 g. Another frequent error is to confuse mass percent with mass by volume percent or to mix units such as grams and kilograms without converting. Careful attention to what the formula asks for, and always adding solute and solvent to obtain total solution mass, avoids these mistakes. Final Answer: The concentration of the solution in terms of mass by mass percentage is 9.35%.

Question 6

In the redox reaction 2PbO + C → 2Pb + CO2, which of the given statements about oxidation and reduction are incorrect?

Accepted Answer

Introduction / Context: This question checks understanding of oxidation and reduction in a typical metal oxide reduction reaction. The example is the reduction of lead oxide with carbon, a common illustration of how metal oxides can be reduced to metals using a reducing agent. Correctly identifying which species are oxidised and reduced is central to mastering redox chemistry, which appears in metal extraction, corrosion, batteries, and many other chemical processes. Given Data / Assumptions: Consider the balanced reaction: 2PbO + C → 2Pb + CO2. Statement A: Carbon dioxide is getting oxidised. Statement B: Lead is getting reduced. Statement C: Carbon is oxidised. Statement D: Lead oxide is getting reduced. The task is to find which statements are incorrect. Concept / Approach: Oxidation and reduction can be understood in terms of oxygen gain and loss or in terms of changes in oxidation number. In metal oxide reductions: The metal oxide usually loses oxygen and is reduced to the metal. The reducing agent, often carbon, gains oxygen and is oxidised to carbon dioxide or carbon monoxide. We assign oxidation numbers to carbon, lead, and oxygen on both sides, then check each statement against these changes to decide whether it is correct or incorrect. Step-by-Step Solution: Step 1: Assign oxidation numbers in 2PbO + C. In PbO: oxygen has oxidation number -2, so lead must be +2 to make the compound neutral. In elemental carbon C: oxidation number is 0. Step 2: Assign oxidation numbers in 2Pb + CO2. In elemental lead Pb: oxidation number is 0. In CO2: oxygen is -2 each, total -4, so carbon must be +4. Step 3: Identify oxidation and reduction. Lead goes from +2 in PbO to 0 in Pb, so lead has been reduced. Carbon goes from 0 in C to +4 in CO2, so carbon has been oxidised. Lead oxide, the compound containing Pb2+, loses oxygen and is reduced to lead metal. Carbon dioxide is a product and does not undergo further change in this reaction, so it is not oxidised here. Verification / Alternative Check: Using the oxygen gain or loss idea, we see that the oxide PbO loses oxygen to form Pb, which is reduction. The element carbon gains oxygen to form CO2, so it is oxidised. Carbon dioxide already contains oxygen and is not gaining still more oxygen in this reaction, so it cannot be described as getting oxidised. This alternative view confirms the result obtained from oxidation numbers and supports our classification of the statements. Why Other Options Are Wrong: Statement A: It claims that carbon dioxide is getting oxidised. In fact, CO2 is a final product and does not further gain oxygen in this reaction, so this statement is incorrect. Statement B: It says lead is getting reduced. Lead metal is the product that results from reduction of lead oxide. However the wording is ambiguous because the species that is reduced is really lead oxide. As per standard teaching, we say PbO is reduced to Pb, not that Pb itself is reduced, so this statement is treated as incorrect. Statement C: It states that carbon is oxidised. This is correct because carbon goes from oxidation number 0 to +4 in CO2. Statement D: It says lead oxide is reduced. This is correct because PbO loses oxygen to form Pb metal. Because A and B are the only incorrect statements, the option that lists A and B together is correct. Common Pitfalls: Learners often confuse the species that is reduced with the metal produced and may say that lead is reduced rather than lead oxide. Another common misunderstanding is to think that any compound containing oxygen is always oxidised further in every reaction, which leads to the mistake about carbon dioxide. Careful use of oxidation numbers and checking which substance gains or loses oxygen helps avoid these errors. Final Answer: The incorrect statements are A and B, so the correct choice is A and B.

Question 7

Metals of very high reactivity are usually extracted from their purified molten ores by which process?

Accepted Answer

Introduction / Context: This question focuses on the metallurgical principle used to extract metals that are highly reactive, such as sodium, potassium, calcium, and aluminium. These metals occupy the top part of the reactivity series and have very strong attraction for oxygen and other non metals. Therefore, ordinary chemical reduction methods are not sufficient to free them from their compounds. Understanding which extraction method is suitable for different reactivity levels is a core concept in basic metallurgy. Given Data / Assumptions: The metals are described as being of high reactivity. The ores are already purified and in molten form. We must choose the process that is realistically used in industry for such metals. Available processes are reduction with a chemical reducing agent, roasting, electrolysis, and calcination. Concept / Approach: The reactivity series tells us that highly reactive metals form very stable compounds. For those metals, conventional reducing agents like carbon cannot reduce the metal oxide or chloride because carbon is less reactive. Instead, we use electrolysis of molten salts or molten ores, where an electric current directly supplies electrons to metal ions at the cathode. This method is used for metals such as sodium from molten sodium chloride and aluminium from molten aluminium oxide dissolved in cryolite. Step-by-Step Solution: Step 1: Identify metals of high reactivity. Examples include K, Na, Ca, Mg, and Al. Step 2: Recall their extraction method. Sodium is obtained by electrolysis of molten sodium chloride. Aluminium is obtained by electrolysis of molten alumina in cryolite (Hall Heroult process). Step 3: Compare with methods given in the options. Reduction by a chemical agent works for less reactive metals like iron and zinc. Roasting and calcination are mainly for converting ores to oxides and for removing volatile impurities, not for extracting highly reactive metals directly. Step 4: Conclude that electrolysis of molten ore is the standard industrial method. Verification / Alternative Check: Looking at standard chemistry textbooks, extraction of sodium and potassium is always shown through Down cell electrolysis of molten chlorides. Aluminium extraction is also always linked to the Hall Heroult electrolytic process. None of these processes rely on carbon reduction at high temperature. Therefore the choice of electrolysis is consistent with the well known industrial practice for highly reactive metals. Why Other Options Are Wrong: Option A: Reduction by an appropriate reducing agent works for moderately reactive metals such as iron from iron oxide using carbon or carbon monoxide. It is not effective for very reactive metals. Option B: Roasting is heating an ore in the presence of oxygen, usually to convert sulphide ores to oxides. It is a preliminary step, not the final extraction method for highly reactive metals. Option D: Calcination is heating an ore in the absence or limited supply of air to remove moisture or volatile impurities, commonly used for carbonate ores. It again is a preparatory step, not the main extraction technique for very reactive metals. Common Pitfalls: A frequent mistake is to think that roasting or calcination always gives the metal itself. In fact, these processes mostly produce oxides which then need further reduction. Another confusion is between extraction of iron, which uses a blast furnace, and extraction of aluminium or sodium, which uses electrolysis. Keeping clear the link between position in the reactivity series and extraction method prevents such confusion. Final Answer: Metals of high reactivity are extracted from purified molten ore by Electrolysis.

Question 8

According to Newlands law of octaves, which of the following was the last element in his classification?

Accepted Answer

Introduction / Context: This question belongs to the history of the periodic table and specifically to Newlands law of octaves. Before the modern long form periodic table was developed, several scientists tried to arrange elements in order of their properties. John Newlands suggested that when elements are arranged in order of increasing atomic mass, every eighth element has similar properties, similar to musical octaves. Knowing which elements he managed to include gives insight into the limitations of his scheme. Given Data / Assumptions: We are dealing with Newlands classification based on the law of octaves. Four candidate elements are listed: rhodium, radium, iron, and thorium. We need to identify which element was the last included in his sequence. Assume typical school level historical description of Newlands work. Concept / Approach: Newlands arranged known elements of his time in increasing order of atomic mass and noticed that every eighth element had similar properties. However his law worked only reasonably well up to a certain point because many elements were not yet discovered and heavier elements did not follow his pattern. Historical records and common exam references mention that Newlands arrangement extended up to thorium. Elements beyond this were either unknown or did not fit his scheme, which contributed to criticism of his work. Step-by-Step Solution: Step 1: Recall the range of elements known to Newlands. When Newlands proposed his law, only about 56 elements were known. Step 2: Recognise from standard teaching that his table was extended only up to thorium. Many textbooks state that thorium was the last element included in Newlands octaves arrangement. Step 3: Compare with the options provided. Rhodium, radium, and iron are not recorded as the terminal element in his arrangement; iron appears earlier in the series. Thorium appears at the end of the sequence he managed to classify under his law. Verification / Alternative Check: The correctness can be cross checked from standard history of chemistry notes, which usually mention that Newlands tried to fit known elements into his octave pattern until thorium. Beyond thorium, the regularity broke down badly, and he did not have space for future discoveries, which was a major limitation of his classification. No standard reference mentions rhodium or radium as the last element in his octave series, reinforcing that thorium is the expected answer in typical exam questions. Why Other Options Are Wrong: Option A: Rhodium is a transition metal that was not highlighted as the last element under Newlands law and does not appear as the terminal element in standard discussions of his table. Option B: Radium is an element that was discovered much later than Newlands time and therefore could not have been part of his classification. Option C: Iron is found earlier among the elements and is not treated as the final element of the octave pattern. Option D: Thorium is widely stated as the last element considered in Newlands octaves, so this is the correct choice. Common Pitfalls: One frequent confusion is between Newlands law of octaves and Mendeleev periodic table. Students may also mix up which scientist dealt with which range of elements. Another mistake is to guess based on atomic numbers or metallic character instead of remembering the historical fact. Keeping a short timeline of Dobereiner triads, Newlands octaves, and Mendeleev periodic law helps keep these contributions distinct in memory. Final Answer: The last element in Newlands law of octaves classification was Thorium.

Question 9

In simple school level chemistry, rusting of iron is best described as which type of chemical change?

Accepted Answer

Introduction / Context: This question examines understanding of rusting, a very familiar example of corrosion. Students commonly learn that rusting involves iron reacting with oxygen and moisture from the air. At basic level it is introduced as oxidation, but more detailed study shows that both oxidation and reduction processes occur, making rusting a redox reaction. Recognising this dual nature is important when moving from simple descriptive chemistry to more rigorous electrochemical thinking. Given Data / Assumptions: The process under discussion is rusting of iron. Rust is mainly hydrated iron(III) oxide formed by reaction with oxygen and water. The options include electrolysis, oxidation, redox reaction, and reduction. We are asked which description best fits rusting. Concept / Approach: A redox reaction is defined as one in which oxidation and reduction occur simultaneously. In rusting, iron atoms lose electrons and are oxidised to Fe2+ and Fe3+ ions, while oxygen molecules gain electrons and are reduced to oxide or hydroxide ions. Thus both oxidation and reduction processes happen together. Although it is not electrolysis in the strict sense, rusting can be viewed as a slow electrochemical process involving local galvanic cells on the metal surface. Step-by-Step Solution: Step 1: Consider what happens to iron. Iron metal (Fe) loses electrons and forms Fe2+ or Fe3+ ions, so iron is oxidised. Step 2: Consider what happens to oxygen. Molecular oxygen from air gains electrons and is reduced to oxide or hydroxide ions in the presence of water. Step 3: Combine these processes. Because oxidation of iron and reduction of oxygen take place simultaneously, the overall process is a redox reaction. Step 4: Relate this to the options. Calling rusting just oxidation is incomplete, because it ignores the reduction step. Calling it reduction is entirely wrong, because the metal loses electrons. Electrolysis is a forced process driven by an external power source, which does not match natural rusting. Verification / Alternative Check: In the electrochemistry chapter, rusting is often explained using small anodic and cathodic regions on the surface of iron. At the anode, iron is oxidised, and at the cathode, oxygen is reduced. The flow of electrons between these regions and the formation of hydroxide ions make it clear that both oxidation and reduction occur together. This confirms that rusting must be treated as a redox process rather than only one half of it. Why Other Options Are Wrong: Option A: Electrolysis requires an external electric current to drive a non spontaneous reaction. Rusting is a spontaneous corrosion process and not an example of deliberate electrolysis. Option B: Oxidation does occur in rusting, but the reaction also has a reduction step, so describing it only as oxidation is incomplete for examination purposes. Option D: Reduction means gain of electrons, but the main metal iron actually loses electrons, so calling rusting a reduction process is incorrect. Common Pitfalls: Many learners memorise early that rusting is oxidation and continue to tick that option even after learning about redox. It is important to update understanding as new concepts are added. Another pitfall is to assume that any reaction involving oxygen is automatically labelled oxidation only, without thinking about the reduction partner. Always remember that real chemical reactions involve both halves of the redox process. Final Answer: Rusting of iron is best described as a Redox reaction (Oxidation and Reduction).

Question 10

Which gas in the atmosphere contributes the most to the Greenhouse effect on Earth?

Accepted Answer

Introduction / Context: Global warming and climate change topics often mention greenhouse gases. Students are usually familiar with carbon dioxide, but in terms of natural greenhouse effect, water vapour actually contributes the largest share. This question checks awareness of the relative roles of different atmospheric gases in trapping heat near the surface of Earth. Given Data / Assumptions: The gases listed are water vapour, ozone, oxygen, and nitrogen. We are concerned with contribution to the greenhouse effect, not just abundance in the atmosphere. Greenhouse effect refers to trapping of outgoing infrared radiation by certain gases. Concept / Approach: Greenhouse gases are substances that strongly absorb infrared radiation emitted by Earth surface and atmosphere. While carbon dioxide, methane, nitrous oxide, and ozone are significant contributors, naturally occurring water vapour accounts for the largest part of the total greenhouse effect. Oxygen and nitrogen dominate by volume in the atmosphere but do not absorb infrared radiation efficiently. So we must identify which gas among the options behaves as a major greenhouse gas and contributes most in practice. Step-by-Step Solution: Step 1: Recall which gases are known greenhouse gases. Water vapour, carbon dioxide, methane, nitrous oxide, and ozone are greenhouse gases. Step 2: Compare relative contributions. Water vapour alone is responsible for a large fraction of the natural greenhouse effect, often quoted as more than half. Carbon dioxide and other gases contribute the rest but are not listed among the options here. Step 3: Analyse options. Ozone does contribute but in a smaller overall share compared to water vapour. Oxygen and nitrogen have almost no greenhouse activity. Thus the gas that contributes the most is water vapour. Verification / Alternative Check: Climate science sources commonly state that if all greenhouse gases were removed, Earth would be much colder and that water vapour accounts for the largest share of the natural greenhouse effect. However human activity mainly alters the concentration of gases like carbon dioxide and methane, which then indirectly affect water vapour levels. Since the question does not ask which gas is the most important anthropogenic greenhouse gas but which gas contributes most overall, water vapour remains the correct answer. Why Other Options Are Wrong: Option B: Ozone absorbs ultraviolet and some infrared radiation and is a greenhouse gas, but its total contribution to the greenhouse effect is smaller than that of water vapour. Option C: Oxygen is very poor at absorbing infrared radiation and is not considered a greenhouse gas in normal discussions. Option D: Nitrogen makes up most of the atmosphere by volume but does not contribute significantly to the greenhouse effect because it hardly absorbs infrared radiation. Common Pitfalls: Because media coverage focuses heavily on carbon dioxide, students may assume it is always the largest contributor in every sense. It is true that carbon dioxide is the most important greenhouse gas to control for climate policy, but that does not change the fact that, under natural conditions, water vapour provides the largest share of greenhouse warming. Carefully reading what the question specifically asks helps avoid this confusion. Final Answer: The gas that contributes most to the greenhouse effect is Water vapour.

Question 11

Constantan is an alloy formed mainly from which combination of metals?

Accepted Answer

Introduction / Context: This question tests knowledge of common engineering alloys. Constantan is a special alloy known for having nearly constant electrical resistance over a range of temperatures. Such alloys are widely used in resistance wires and thermocouples. Knowing its composition helps students connect material properties with alloying elements in basic materials science and chemistry. Given Data / Assumptions: The alloy in question is constantan. Options list different combinations of metals such as copper, nickel, manganese, chromium, iron, and titanium. We assume standard textbook composition where constantan is mainly copper and nickel. Concept / Approach: Alloys are mixtures of metals designed to obtain certain properties. Constantan is particularly valued because its resistivity is almost unchanged with temperature. The classic constantan alloy contains about 55 percent copper and 45 percent nickel by mass. While small impurities may exist, these two metals are the primary components. Therefore, among the given options, the pair containing only copper and nickel must be identified. Step-by-Step Solution: Step 1: Recall the definition of constantan. Constantan is a copper nickel alloy with almost constant resistivity. Step 2: Scan the options for a copper nickel pair. Option B includes Cu, Ni, and Mn. Option D includes Cu and Ni only. Step 3: Determine which matches the standard composition. Standard constantan is primarily copper and nickel, not a four metal or three metal mixture. Therefore, the correct combination is Cu and Ni. Verification / Alternative Check: Reference tables for resistance alloys list constantan as a copper nickel alloy often described by formulas like Cu55Ni45. Some grades may include minor elements but the core remains that constantan is basically a copper nickel system. There is no conventional constantan alloy described by compositions such as Ni Cr Mn Fe, which are more typical of stainless steels or nichrome type alloys, confirming that Cu, Ni is the correct combination. Why Other Options Are Wrong: Option A: The combination Ni, Ti, Fe, Cr resembles high strength or stainless alloys rather than constantan and is not the textbook composition. Option B: Cu, Ni, Mn adds manganese, which is part of some other special alloys but is not characteristic of constantan in basic chemistry courses. Option C: Ni, Cr, Mn, Fe suggests stainless steel or heat resistant alloys, not constantan. Common Pitfalls: Students sometimes confuse constantan with nichrome or stainless steel, all of which are used as heating elements or resistance materials. Remember that nichrome is based on nickel and chromium, stainless steels are iron based with chromium and sometimes nickel, whereas constantan is specifically a copper nickel alloy. Associating each name with its key metals is a good memory aid. Final Answer: Constantan is mainly an alloy of Cu, Ni, that is copper and nickel.

Question 12

For a standard hydrogen electrode, which statement about its electrode potentials is correct?

Accepted Answer

Introduction / Context: This question concerns the standard hydrogen electrode, often abbreviated as SHE, which is the reference electrode used to define standard electrode potentials. In electrochemistry, all standard reduction potentials are measured relative to this electrode. It is assigned a conventional value, but its true absolute potential in volts is different from that conventional assignment. The question tests whether the learner can distinguish between absolute and conventional standard potentials. Given Data / Assumptions: Standard hydrogen electrode is considered under standard conditions. Four statements about its potentials are given. Standard electrode potential of SHE is conventionally taken as zero volt. Absolute potential is the potential of the electrode relative to a vacuum reference, which is more difficult to measure. Concept / Approach: By international convention, the standard electrode potential of the hydrogen half cell, written as 2H+ + 2e− → H2, is assigned as exactly 0.00 volt at all temperatures where standard conditions apply. This is not because the true potential is physically zero but because we need a reference point. In reality, when compared to a theoretical vacuum level, the hydrogen electrode has a non zero absolute potential, often quoted around 4.4 volts. Therefore, any statement claiming that the absolute electrode potential is zero is incorrect, while the statement that only the absolute potential is non zero is correct. Step-by-Step Solution: Step 1: Recall the convention. The standard hydrogen electrode is assigned a standard electrode potential E standard = 0.00 volt by definition. Step 2: Understand absolute potential. Absolute electrode potential is measured relative to a vacuum and is not set by arbitrary convention. Experimental and theoretical studies show that this absolute value is around 4.4 volts, not zero. Step 3: Evaluate each statement. Statement A: Absolute electrode potential is not zero. This agrees with the idea that the true physical potential has a non zero value. Statement B: Absolute electrode potential is zero. This contradicts the above understanding. Statement C: Both absolute and standard electrode potentials are zero. This cannot be correct because only the standard value is defined as zero. Statement D: Electrode potential is zero only at 25 degree Celsius. In fact the standard value is taken as zero at all temperatures under standard conditions, so this is also wrong. Verification / Alternative Check: Electrochemistry tables explicitly state that the standard hydrogen electrode has been assigned a standard reduction potential of zero volt and that all other electrode potentials are measured relative to this. Specialist sources on absolute electrode potentials mention approximate values around 4.4 volts. Since the question contrasts absolute and standard potentials, we can confidently say that the absolute one is not zero. This cross check confirms statement A as the correct one. Why Other Options Are Wrong: Option B: Saying that the absolute electrode potential is zero ignores the physical meaning of absolute potential and is inconsistent with theoretical and experimental estimates. Option C: Claiming both absolute and standard potentials are zero combines the errors of option B and contradicts known data. Option D: The standard hydrogen electrode is defined so that its standard potential remains zero by convention at all temperatures, not only at 25 degree Celsius, so limiting the statement to a single temperature is incorrect. Common Pitfalls: Students sometimes treat the assigned standard value and true physical value as the same quantity. It is important to remember that standard electrode potentials are relative numbers based on a chosen reference. Confusing absolute potential with standard potential can lead to wrong answers in conceptual questions like this. Keeping the distinction clear between a conventional zero point and a measured quantity avoids this confusion. Final Answer: For a standard hydrogen electrode, the correct statement is that absolute electrode potential is not zero.

Question 13

Which one of the following elements is essential as the central metal for the formation of chlorophyll in green plants?

Accepted Answer

Introduction / Context: This question links basic chemistry with plant biology. Chlorophyll is the green pigment that enables plants to perform photosynthesis by absorbing light energy. At its centre, the chlorophyll molecule contains a specific metal ion. Recognising which metal is present helps students connect the roles of elements like magnesium and iron in biological systems and in fertilisers. Given Data / Assumptions: The substance under discussion is chlorophyll, the main photosynthetic pigment in green plants. Options are calcium, iron, magnesium, and potassium. We assume standard chlorophyll structure taught at school level. Concept / Approach: Chlorophyll molecules are based on a porphyrin type ring structure similar in some ways to the haem group in haemoglobin. In haemoglobin, iron is the central metal ion, whereas in chlorophyll the central ion is magnesium. This magnesium ion is coordinated by nitrogen atoms in the ring and is critical for absorbing light and transferring energy. Therefore the correct answer is the element whose ions occupy that central position, which is magnesium. Step-by-Step Solution: Step 1: Recall the structure of chlorophyll. Chlorophyll has a large ring system with a metal ion at its centre. Step 2: Identify the central ion. Textbooks describe this ion as Mg2+, indicating magnesium. Step 3: Compare with the given options. Calcium is mainly used for cell wall strength, not as the chlorophyll centre. Iron is central in haem but not in chlorophyll. Potassium has roles in stomatal movement and ionic balance, but again not as the centre of chlorophyll. Thus the correct choice is magnesium. Verification / Alternative Check: Information about plant nutrition notes that magnesium deficiency leads to yellowing of leaves, called chlorosis, because chlorophyll cannot be synthesised properly without magnesium. This is strong evidence that magnesium is essential for chlorophyll formation rather than the other listed elements. Agricultural advice on fertilisers also mentions magnesium containing salts for correcting such deficiencies. Why Other Options Are Wrong: Option A: Calcium is essential for plants but is mainly involved in cell wall structure and membrane stability, not as the centre of chlorophyll. Option B: Iron plays a key role in haemoproteins and electron transport chains, yet it is not the central metal in chlorophyll. Option D: Potassium regulates osmotic balance and enzyme activation but does not occupy the central position in the chlorophyll molecule. Common Pitfalls: Students sometimes confuse magnesium with iron because both elements appear in biology as important micronutrients. The easiest way to remember is to link haemoglobin with iron and chlorophyll with magnesium, two parallel examples of metal containing biomolecules. Mixing up these associations can easily lead to the wrong answer in this type of question. Final Answer: The element essential for chlorophyll formation is Magnesium.

Question 14

In the neutralisation reaction between an acid and a base, Acid + Base → _______ + Water, what product correctly fills the blank?

Accepted Answer

Introduction / Context: This question addresses the basic idea of neutralisation in acid base chemistry. When an acid reacts with a base, they cancel each other effect to form more neutral products. Students learn a general word equation for this process, and correctly identifying the missing product is important for understanding many real world applications, such as antacid use, wastewater treatment, and laboratory titrations. Given Data / Assumptions: General reaction pattern: acid plus base produces something plus water. The options for the missing product are carbon dioxide, metal oxide, hydrogen gas, and salt. We consider typical reactions like hydrochloric acid with sodium hydroxide. Concept / Approach: The classic neutralisation reaction between an acid and a base can be represented in ionic form as H+ plus OH− giving H2O. In molecular examples, the metal part of the base and the non metal part of the acid combine to form a salt. For instance, HCl plus NaOH gives NaCl plus H2O. Therefore the general pattern is acid plus base gives a salt and water. Carbon dioxide is produced in reactions of acids with carbonates, not simple bases. Hydrogen gas is released when acids react with some metals, not when they react with bases. Step-by-Step Solution: Step 1: Write a typical neutralisation example. HCl + NaOH → NaCl + H2O Step 2: Identify the products. The first product is NaCl, a salt, and the second product is water. Step 3: Compare with the word equation given. Acid + Base → _______ + Water. Here the blank must represent the salt that forms from cation of the base and anion of the acid. Step 4: Choose the matching option. Among carbon dioxide, metal oxide, hydrogen gas, and salt, only salt correctly fits the word equation. Verification / Alternative Check: Consider another example: sulphuric acid reacting with sodium hydroxide. The balanced equation is H2SO4 + 2NaOH → Na2SO4 + 2H2O. Again, the products are a salt (sodium sulphate) and water, confirming that salt is the general product. Other combinations, such as hydrochloric acid with magnesium hydroxide or calcium hydroxide, follow the same pattern, reinforcing that the blank should always be filled with salt. Why Other Options Are Wrong: Option A: Carbon dioxide is produced when acids react with carbonates or bicarbonates, for example HCl with CaCO3, not in simple acid base neutralisation. Option B: Metal oxides can behave as bases or acidic oxides but they are reactants or intermediates, not the typical product of mixing acid and base in solution. Option C: Hydrogen gas is evolved when active metals such as zinc or magnesium react with acids, not when a base is present. Common Pitfalls: Learners sometimes mix up different general reaction patterns, such as acid plus carbonate giving salt, carbon dioxide, and water. Confusing those with simple neutralisation leads to wrong choices. A helpful tip is to remember that the word neutralisation emphasises the formation of water from H+ and OH−, with the ions left behind combining into a salt. Final Answer: In the reaction Acid + Base → _______ + Water, the blank is filled by Salt.

Question 15

The pH of healthy human blood is best described as which of the following?

Accepted Answer

Introduction / Context: The pH scale is used to describe how acidic or basic a solution is. Human blood has a tightly regulated pH, and understanding whether it is acidic or basic helps connect chemistry with human physiology. Small deviations from the normal range can have serious health consequences, so this is also an important point in biology and medical entrance examinations. Given Data / Assumptions: Typical healthy arterial blood pH is about 7.35 to 7.45. On the pH scale, 7 is neutral, values below 7 are acidic, and values above 7 are basic or alkaline. The options ask us to choose slightly acidic, highly acidic, slightly basic, or highly basic. Concept / Approach: Since blood pH lies slightly above 7, it must be on the basic side. However, the difference from neutrality is small, so it is not correct to say highly basic. Instead, we classify blood as slightly basic or slightly alkaline. The body maintains this narrow range through buffer systems like bicarbonate, haemoglobin, and phosphate buffers, as well as respiratory and renal regulation. Step-by-Step Solution: Step 1: Recall the normal pH range of human blood. It is approximately 7.35 to 7.45. Step 2: Compare this range with neutral pH. Neutral pH is 7.0, so blood pH is slightly greater than 7. Step 3: Decide on acidity or basicity. Any solution with pH greater than 7 is basic or alkaline, while pH less than 7 would be acidic. Since blood pH is only slightly above 7, it must be slightly basic. Verification / Alternative Check: Medical references define acidosis as a condition where blood pH tends to fall below the normal range, that is, below 7.35, and alkalosis where pH increases above 7.45. Both are considered abnormal, reinforcing that the healthy baseline is a slightly basic region centred around 7.4. This confirms that describing blood as slightly basic is correct, whereas calling it acidic or highly basic does not match observed values. Why Other Options Are Wrong: Option A: Slightly acidic would imply pH just below 7, which is not consistent with measured values of normal blood. Option B: Highly acidic would correspond to much lower pH values, such as 1 or 2, which would be incompatible with life. Option D: Highly basic would imply a pH value far above 7, such as 10 or 12, again impossible for normal physiology. Common Pitfalls: Some learners mistakenly assume that blood must be neutral because it is so important to the body, or they confuse the idea that the stomach is acidic with the condition of blood. Others misremember the pH range and think 7.0 is the exact value. Careful attention to diagrams and tables in textbooks, which usually show blood at pH 7.4, helps avoid such confusion. Final Answer: The pH of human blood is Slightly Basic.

Question 16

Which of the following elements is an inert noble gas?

Accepted Answer

Introduction / Context: Inert gases, also called noble gases, are elements in group 18 of the periodic table that show very low chemical reactivity under normal conditions. They are commonly used in lighting, welding, and as protective atmospheres. This question tests whether the learner can recognise one such gas among familiar atmospheric components. Given Data / Assumptions: Candidates listed are hydrogen, nitrogen, oxygen, and argon. Inert gases are noble gases such as helium, neon, argon, krypton, xenon, and radon. We assume standard periodic table groupings. Concept / Approach: Noble gases have completely filled valence electron shells, making them chemically stable. Argon, with atomic number 18, has a full outer shell and belongs to group 18. In contrast, hydrogen (group 1), nitrogen (group 15), and oxygen (group 16) are reactive non metals that form many compounds. Therefore, we identify which of the options is in group 18 and hence inert. Step-by-Step Solution: Step 1: Recall the list of noble gases. Helium, neon, argon, krypton, xenon, and radon are noble gases. Step 2: Compare this list with the given options. Among hydrogen, nitrogen, oxygen, and argon, only argon appears in the noble gas list. Step 3: Conclude that argon is the inert gas in the given set. Verification / Alternative Check: Looking at the periodic table, argon is in group 18, period 3. Its electron configuration ends with 3s2 3p6, indicating a complete octet. Hydrogen has a single electron and is reactive, nitrogen forms nitrates, ammonia, and other compounds, and oxygen participates in combustion and respiration reactions. Only argon remains largely unreactive under ordinary conditions, confirming our choice. Why Other Options Are Wrong: Option A: Hydrogen is reactive and forms many compounds, such as water and acids. It is not a noble gas. Option B: Nitrogen, although relatively less reactive at room temperature due to a strong triple bond in N2, can form many compounds and is not classified as a noble gas. Option C: Oxygen is a very reactive element that supports combustion and is essential for respiration, not an inert gas. Common Pitfalls: Sometimes students confuse low reactivity with complete inertness and may think nitrogen is inert simply because air is mostly nitrogen and it does not catch fire easily. However, the term inert gas in the periodic table context refers specifically to group 18 noble gases. Keeping this distinction clear helps to avoid mixing up nitrogen and noble gases. Final Answer: The inert noble gas among the options is Argon.

Question 17

Which among the following food items is not an example of an emulsion type colloid?

Accepted Answer

Introduction / Context: This question tests understanding of colloids and emulsions in everyday foods. Many dairy products are emulsions, where tiny droplets of one liquid are dispersed in another immiscible liquid. Recognising which items are emulsions and which are not helps students apply abstract concepts to common substances like milk, butter, cream, and curd. Given Data / Assumptions: Options include chocolate milk, butter, whipped cream, and curd. Emulsion is defined as a colloid in which both dispersed phase and dispersion medium are liquids. Dairy products often involve fat dispersed in water or vice versa. Concept / Approach: Milk is a classic example of an oil in water emulsion, with fat droplets dispersed in water. Chocolate milk is milk with added cocoa, so it still largely behaves as an emulsion. Butter is considered a water in oil emulsion, with tiny water droplets dispersed in fat. Whipped cream originates from cream, which is itself an emulsion, and incorporates air to form a foam but still relies on the underlying emulsion structure. Curd or yoghurt, on the other hand, is formed by coagulation of proteins creating a gel network, not by liquid droplets dispersed in another liquid, so it is better classified as a gel rather than a true emulsion. Step-by-Step Solution: Step 1: Analyse chocolate milk. Milk is an emulsion of fat in water. Adding chocolate powder does not change the basic emulsion nature, so chocolate milk remains a type of emulsion. Step 2: Analyse butter. Butter consists of water droplets dispersed in fat, which is a water in oil emulsion. Step 3: Analyse whipped cream. Whipped cream is formed by whipping cream, an emulsion of fat in water, introducing air to create a foam but still based on an emulsion system. Step 4: Analyse curd. Curd is formed when milk proteins coagulate into a solid network trapping water and fat. This is more accurately described as a gel, not liquid in liquid emulsion. Therefore, curd is the item that is not an example of an emulsion. Verification / Alternative Check: Standard colloid classification lists milk, cream, butter, and mayonnaise as emulsions. Curd is usually not grouped with emulsions but with gels, because its continuous phase is solid like and not simply a liquid dispersion. This classification supports the conclusion that among the given options, curd is the one that does not fit the definition of an emulsion. Why Other Options Are Wrong: Option A: Chocolate milk is milk based and therefore an oil in water emulsion with added suspended cocoa solids. Option B: Butter is a water in oil emulsion where water droplets are dispersed in solidified fat. Option C: Whipped cream is a foam built on an underlying emulsion of fat in water, so it is considered an emulsion derived product. Common Pitfalls: Students may think that curd is similar to milk and thus must also be an emulsion. However, the key difference is that curd formation involves coagulation of casein proteins into a semi solid framework, making it a gel. Another pitfall is to ignore the precise definition of an emulsion and group all dairy products together without considering their microstructure. Final Answer: The food item that is not an example of an emulsion is Curd.

Question 18

Which of the following processes is a physical change rather than a chemical change?

Accepted Answer

Introduction / Context: This question checks whether the learner can distinguish between physical and chemical changes. A physical change does not produce a new substance and is usually reversible by simple physical means. A chemical change leads to new substances with different properties, often accompanied by colour change, gas evolution, or energy release. Recognising these differences is a foundational skill in elementary chemistry. Given Data / Assumptions: Four processes are listed: burning coal, burning wood, heating a platinum crucible, and heating potassium chlorate. We must identify the one that is purely physical. Chemical changes generally involve chemical reactions, while physical changes involve changes of state, shape, or size without altering composition. Concept / Approach: Burning processes, such as combustion of coal or wood, clearly involve chemical reactions with oxygen and produce new substances like carbon dioxide, water vapour, and ash. Heating of potassium chlorate decomposes the compound to potassium chloride and oxygen gas, which again is a chemical change. Heating a platinum crucible mainly causes expansion due to temperature rise and possibly glowing, but platinum is chemically very inert and does not react under ordinary heating, so no new substance is formed. This makes it a physical change. Step-by-Step Solution: Step 1: Examine burning of coal. Coal reacts with oxygen to form carbon dioxide, carbon monoxide, and ash, which are new substances, so this is a chemical change. Step 2: Examine burning of wood. Wood combustion produces carbon dioxide, water vapour, and ash, again a chemical change. Step 3: Examine heating of a platinum crucible. Platinum is a noble metal that resists oxidation. On heating, it may glow red or white but retains its chemical identity as metallic platinum, so this is a physical change. Step 4: Examine heating of potassium chlorate. Potassium chlorate decomposes on heating into potassium chloride and oxygen gas, clearly a chemical reaction. Verification / Alternative Check: The easiest way to verify is to ask whether a new chemical substance is produced. Both burning examples and the decomposition of potassium chlorate produce new substances. In contrast, heating a platinum crucible simply changes its temperature and colour, and the metal returns to its original state on cooling. Its mass and composition remain unchanged, confirming that this is a physical change. Why Other Options Are Wrong: Option A: Burning coal is combustion of carbon based fuel, which is a chemical process. Option B: Burning wood involves complex chemical reactions of cellulose, lignin, and other components with oxygen, again a chemical change. Option D: Heating potassium chlorate leads to thermal decomposition, forming new substances, which is a chemical reaction. Common Pitfalls: A common mistake is to assume that any process involving heat is a chemical change. Learners must remember that heat can cause both physical phenomena like expansion and state changes, and also chemical reactions, depending on the substance and conditions. Another pitfall is not knowing that platinum is highly inert, so they may mistakenly think it oxidises easily when heated, which is not the case in normal laboratory conditions. Final Answer: The process that represents a physical change is Heating of a platinum crucible.

Question 19

Calcium (atomic number 20) and argon (atomic number 18) both have mass number 40. Such pairs of nuclides are called what?

Accepted Answer

Introduction / Context: This question belongs to nuclear chemistry and the classification of nuclides. It involves comparing atomic numbers and mass numbers of different elements. Several terms like isotopes, isobars, and isotones are used to describe relationships between nuclides with regard to protons, neutrons, and total nucleons. Correct usage of these terms is very common in objective type exams. Given Data / Assumptions: Calcium has atomic number 20 and mass number 40. Argon has atomic number 18 and mass number 40. Therefore, both nuclides have the same mass number but different atomic numbers. Options include isotones, isochores, isobars, and isotopes. Concept / Approach: The definitions are as follows: Isotopes: same atomic number, different mass numbers. Isobars: same mass number, different atomic numbers. Isotones: same number of neutrons, different atomic numbers. Isochores is not commonly used in this nuclear sense and may refer to constant volume lines in thermodynamics. Since calcium 40 and argon 40 have the same mass number 40 but different atomic numbers 20 and 18, they clearly fit the definition of isobars. Step-by-Step Solution: Step 1: Note the mass numbers. Calcium nuclide given has mass number 40. Argon nuclide given also has mass number 40. Step 2: Note the atomic numbers. Calcium has atomic number 20, argon has 18, so they are different elements. Step 3: Match with definitions. Same mass number but different atomic numbers defines isobars. Therefore calcium 40 and argon 40 are isobars. Verification / Alternative Check: We can compute the number of neutrons to cross check. Calcium 40 has 40 minus 20 equals 20 neutrons. Argon 40 has 40 minus 18 equals 22 neutrons. Since the number of neutrons is not the same, they are not isotones. Because they have different atomic numbers, they cannot be isotopes of the same element either. The only correct classification based on same mass number is isobars. Why Other Options Are Wrong: Option A: Isotones would require equal number of neutrons, which is not true here. Option B: Isochores relate to constant volume processes in physical chemistry and do not describe a relationship between nuclides. Option D: Isotopes have the same atomic number but different mass numbers, which is not the case because calcium and argon are different elements. Common Pitfalls: Learners often confuse isobars and isotopes because both involve the word bar or tope and both relate to mass and atomic numbers. A good memory aid is that isotopes share the same atomic number (same element), while isobars share the same mass number but are different elements. Writing small tables of Z, A, and N for a few examples can help fix these concepts firmly. Final Answer: Calcium 40 and argon 40 are isobars of each other.

Question 20

What is the number of atoms present in 46 g of sodium 23, expressed in terms of Avogadro constant N?

Accepted Answer

Introduction / Context: This question is a straightforward application of the mole concept. It asks how many atoms of an element are present in a given mass when the molar mass and Avogadro constant are known. Such questions are foundational for stoichiometry, limiting reagent problems, and understanding chemical equations quantitatively. Given Data / Assumptions: Element considered is sodium 23, meaning its atomic mass is 23 g per mole. Given mass of sodium is 46 g. Avogadro constant is represented by N, which is the number of particles in 1 mole. We assume sodium is in atomic form, so each particle is an atom. Concept / Approach: Number of moles is calculated by dividing the given mass by molar mass. Once the number of moles is known, the total number of atoms is obtained by multiplying the number of moles by Avogadro constant. In symbols: moles = given mass / molar mass number of atoms = moles * N We then simplify this expression and match it with the symbolic options given (N/2, N, 2N, 23 N). Step-by-Step Solution: Step 1: Calculate the number of moles of sodium. molar mass of sodium 23 = 23 g per mole. given mass = 46 g. moles of sodium = 46 / 23 = 2 moles. Step 2: Find the number of atoms. Each mole contains N atoms. Therefore, number of atoms in 2 moles = 2 * N = 2N. Step 3: Match the result with the options. 2N corresponds exactly to option C. Verification / Alternative Check: As a quick sanity check, if 23 g of sodium contains N atoms, then doubling the mass to 46 g while keeping the same substance should double the number of atoms. So the answer must be twice N, that is 2N. This proportional reasoning confirms the result without needing to redo the full calculation. Why Other Options Are Wrong: Option A: N/2 would correspond to half a mole, which would be 11.5 g of sodium, not 46 g. Option B: N corresponds to one mole, which is 23 g, again not matching 46 g. Option D: 23 N would be 23 times Avogadro constant, which would require 23 moles of sodium, that is 23 * 23 = 529 g, far more than 46 g. Common Pitfalls: Misreading the question and treating 23 as a multiplier for N instead of as the molar mass is a common mistake. Another error is to forget to divide the mass by molar mass before multiplying by N, leading to expressions like 46 N instead of 2N. Carefully keeping the steps in the right order prevents such mistakes. Final Answer: The number of atoms in 46 g of sodium 23 is 2N.

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