Question 1

Definition recall: Acidity or alkalinity (basicity) of an aqueous solution is expressed by pH, defined in terms of hydrogen-ion concentration [H+]. Which expression is correct?

Accepted Answer

Introduction / Context: pH is ubiquitous in chemical process control, water treatment, biochemical engineering, and environmental monitoring. A precise definition avoids significant control errors in neutralization and fermentation. Given Data / Assumptions: [H+] denotes molar concentration (or, more rigorously, activity) of hydrogen ions in aqueous solution at the measurement temperature. Base-10 logarithm is implied in the conventional pH scale. Concept / Approach: The Sorensen definition is pH = −log10(a_H+). For ideal dilute solutions, activity a_H+ ≈ [H+], giving pH = −log10[H+]. Low [H+] (acidic) yields higher log magnitude but with a negative sign, resulting in small pH values; high [H+] reduces the pH further. Step-by-Step Solution: Start from the definition: pH = −log10(a_H+).Assume ideality: a_H+ ≈ [H+].Thus, pH = −log10[H+]. Verification / Alternative check: At neutrality (25°C), [H+] ≈ 1.0 × 10^-7 M → pH = −log10(10^-7) = 7, which matches the standard neutral pH reference. Why Other Options Are Wrong: pH = log10[H+] lacks the minus sign and reverses acidity–basicity interpretation; 1/[H+] is dimensionally inconsistent; −ln[H+] uses natural log and would not align with the conventional scale. Common Pitfalls: Confusing log bases; ignoring activity corrections in high-ionic-strength solutions. Final Answer: pH = -log10[H+]

Question 2

Empirical formula from mass data: A compound contains nitrogen and oxygen in masses of 28 g and 80 g, respectively. What is its formula?

Accepted Answer

Introduction / Context: Deriving a chemical formula from elemental mass data is a basic but essential skill for process chemists and engineers. The method uses molar masses to convert mass to moles, then simplifies to the smallest whole-number ratio. Given Data / Assumptions: Mass of nitrogen = 28 g; atomic mass N ≈ 14 g/mol. Mass of oxygen = 80 g; atomic mass O ≈ 16 g/mol. Assume the given masses correspond to one formula unit ratio (empirical composition). Concept / Approach: Compute moles of each element, divide by the smallest mole count to obtain a whole-number ratio, and map to the simplest empirical formula. If the empirical formula matches a known molecular formula under typical valences, it is likely the molecular formula as well (nitrogen pentoxide is a well-known compound). Step-by-Step Solution: n_N = 28 g / 14 g·mol^-1 = 2.00 mol.n_O = 80 g / 16 g·mol^-1 = 5.00 mol.Mole ratio N:O = 2:5 → empirical formula N2O5. Verification / Alternative check: Check valence consistency: nitrogen can reach +5 in N2O5; oxygen is −2, satisfying charge neutrality with known oxide N2O5. Why Other Options Are Wrong: N2O4 implies a 1:2 N:O ratio; N2O3 implies 2:3; NO2 is 1:2; none match 2:5. Common Pitfalls: Forgetting to reduce to the simplest whole-number ratio; rounding errors when mole ratios are near halves or thirds. Final Answer: N2O5

Question 3

Limits of Raoult’s law: Raoult’s law is not applicable under which conditions?

Accepted Answer

Introduction / Context: Raoult’s law is the foundation for modeling the vapor pressure of ideal solutions, but practitioners must know its limits to avoid large design errors in distillation and evaporation. Deviations are common when molecular interactions differ significantly from like–like interactions or when composition moves far from dilute or ideal ranges. Given Data / Assumptions: We seek conditions where Raoult’s law fails or becomes inaccurate. Focus on liquid-phase nonideality due to association/dissociation and high concentrations. Concept / Approach: Raoult’s law assumes activity coefficients γ_i ≈ 1. Association (e.g., carboxylic acid dimerization) or dissociation (e.g., electrolytes) changes effective species counts and interactions, leading to γ_i ≠ 1. At higher concentrations, many solutions exhibit strong nonideality due to specific interactions, size disparity, or hydrogen bonding, again violating Raoult’s assumptions. In contrast, dilute solutions of non-volatile solutes used for colligative properties often approximate Raoult’s behavior for the solvent. Step-by-Step Solution: Identify non-ideal mechanisms: association/dissociation → non-unity activity coefficients.Recognize composition effect: concentrated solutions magnify deviations.Therefore, conditions (a) and (b) are where Raoult’s law is not applicable. Verification / Alternative check: Electrolyte solutions require activity coefficient models (e.g., Debye–Hückel, Pitzer). Strongly non-ideal nonelectrolytes use Margules, NRTL, or UNIQUAC—not Raoult’s law. Why Other Options Are Wrong: Low-concentration non-volatile solute in a similar solvent often follows Raoult’s law for the solvent; an ideal binary of similar molecules is precisely the scenario where Raoult’s law performs well. Common Pitfalls: Applying Raoult’s law to electrolytes or associating systems; ignoring temperature/composition dependence of γ_i. Final Answer: both (a) and (b)

Question 4

At NTP (0 °C and 1 atm), 2 litres of nitrogen gas (N2) are contained in a vessel. Approximately how many grams does this amount of nitrogen weigh?

Accepted Answer

Introduction / Context: NTP (normal temperature and pressure) conditions simplify gas calculations. At NTP (taken here as 0 °C and 1 atm), one mole of an ideal gas occupies about 22.4 litres. This problem checks understanding of molar volume and mass–mole conversions for nitrogen, N2, which is common in chemical engineering material balance work. Given Data / Assumptions: NTP ≈ 0 °C (273 K) and 1 atm total pressure. Molar volume at NTP ≈ 22.4 L per mol (ideal gas approximation). Molar mass of N2 = 28 g per mol. Gas behaves ideally; container effects are negligible. Concept / Approach: The number of moles is computed using the molar volume at NTP. Once moles are known, multiply by the molar mass to obtain mass in grams. The linear proportionality between volume and moles at fixed T and P is the key idea. Step-by-Step Solution: Moles of N2, n = V / (molar volume) = 2 L / 22.4 L·mol^-1.n ≈ 0.08929 mol.Mass = n * MW = 0.08929 mol * 28 g·mol^-1.Mass ≈ 2.5 g (to two significant figures). Verification / Alternative check: As a sense check, 22.4 L of N2 would weigh 28 g at NTP; therefore 2 L is about 2/22.4 of 28 g, which is 2.5 g. The ratio method confirms the calculation without a calculator. Why Other Options Are Wrong: 14 and 28 g: These are the masses for 0.5 mol and 1 mol of N2, respectively, not for 2 L. 1.25 g: This would correspond to 1 L of N2 at NTP (half of 2.5 g). Common Pitfalls: Using the molar mass of atomic nitrogen (14 g per mol) instead of diatomic nitrogen (28 g per mol), or using the wrong molar volume (e.g., 24.0 L per mol at 25 °C) can lead to errors. Final Answer: 2.5

Question 5

A liquid has density 1500 kg/m^3. Convert this value to grams per litre (g/L). What is the correct numerical value?

Accepted Answer

Introduction / Context: Unit conversions between SI base units and practical laboratory units are essential in process calculations. Converting kg/m^3 to g/L is common in solutions, density measurements, and material balances. Given Data / Assumptions: Density ρ = 1500 kg/m^3. 1 m^3 = 1000 L. 1 kg = 1000 g. Concept / Approach: Convert the mass unit from kilograms to grams and the volume unit from cubic metres to litres. Because both conversions involve a factor of 1000, the numerical value is unchanged, but the unit switches to g/L. Always track units carefully to avoid mistakes. Step-by-Step Solution: ρ (g/L) = ρ (kg/m^3) * (1000 g / 1 kg) * (1 m^3 / 1000 L)ρ (g/L) = 1500 * (1000 / 1000) = 1500 g/LThus, the density is 1500 g/L. Verification / Alternative check: You can reason dimensionally: kg/m^3 to g/L has a 1:1 numeric mapping since both numerator and denominator scale by 1000. Hence, 1500 kg/m^3 ↔ 1500 g/L directly. Why Other Options Are Wrong: 1.5, 15, 150: These miss one or two powers of ten and reflect common slip-ups in thousand-fold conversions. Common Pitfalls: Forgetting that 1 m^3 is 1000 L (not 100 L), or mixing up g/mL with g/L. Also, do not confuse specific gravity (relative to water) with absolute density. Final Answer: 1500

Question 6

Identify the correct unit(s) of dynamic viscosity (absolute viscosity). Which of the following correctly represents dynamic viscosity in the CGS system?

Accepted Answer

Introduction / Context: Viscosity units are frequently encountered in fluid mechanics, rheology, and transport phenomena. Distinguishing dynamic viscosity (μ) from kinematic viscosity (ν) and matching the correct unit system prevents serious calculation errors. Given Data / Assumptions: Dynamic viscosity μ measures shear stress per unit velocity gradient. CGS units are commonly used alongside SI in older literature. We compare the options with known unit definitions. Concept / Approach: In CGS, dynamic viscosity is measured in poise (P), where 1 P = 1 g/(cm·s). Kinematic viscosity is measured in stokes (St), where 1 St = 1 cm^2/s. In SI, μ uses Pa·s or N·s/m^2 equivalently. Step-by-Step Solution: Dynamic viscosity (CGS): poise = g/(cm·s).Kinematic viscosity (CGS): stoke = cm^2/s.Therefore, “poise” and “gm/cm·s” are both correct for dynamic viscosity. Verification / Alternative check: Convert between SI and CGS: 1 Pa·s = 10 P. This cross-check helps align SI values with CGS references when required. Why Other Options Are Wrong: Stoke: This is for kinematic viscosity, not dynamic viscosity. Common Pitfalls: Confusing ν with μ; remember ν = μ/ρ. Using the wrong property unit in Reynolds number or pump power calculations can lead to major design errors. Final Answer: both (b) & (c)

Question 7

Specific heat at constant pressure (Cp): Choose the correct SI unit for Cp used in thermodynamics and heat transfer.

Accepted Answer

Introduction / Context: The unit of specific heat capacity at constant pressure (Cp) is fundamental to energy balances, heat exchanger design, and process simulations. Correct unit usage ensures consistent calculations of enthalpy changes and sensible heat loads. Given Data / Assumptions: Sought: SI unit for Cp. Cp relates heat input to temperature change per unit mass at constant pressure. Concept / Approach: By definition, Cp = (heat added at constant P) / (mass * temperature rise). In SI, heat is measured in joules (J), mass in kilograms (kg), and temperature in kelvin (K). Therefore, Cp units are J/(kg·K). Step-by-Step Solution: Cp = Q / (m * ΔT) at constant pressure.Units = J / (kg * K) = J/kg·K.Hence, the correct SI unit is J/kg·K. Verification / Alternative check: Enthalpy change for sensible heating: Δh = ∫ Cp dT, which has units J/kg—consistent only if Cp is J/kg·K integrated over K. Why Other Options Are Wrong: W/m^2·°C and W/m·K: These are heat flux or thermal conductivity units, not specific heat. J/m^3·K: This is volumetric heat capacity, not specific heat per unit mass. Common Pitfalls: Mixing mass-based Cp with molar heat capacity (J/mol·K). Always confirm whether a property is mass-based or molar-based. Final Answer: J/kg·K

Question 8

For a monatomic ideal gas, what is the ratio of specific heats (Cp/Cv), commonly denoted as gamma?

Accepted Answer

Introduction / Context: The ratio of specific heats γ = Cp/Cv influences speed of sound, isentropic relations, and compressor/expander performance. For monatomic gases, γ takes a characteristic value based on degrees of freedom from kinetic theory. Given Data / Assumptions: Monatomic ideal gas (e.g., He, Ne, Ar). Equipartition of energy applies. Concept / Approach: For a monatomic gas, translational degrees of freedom = 3. Then Cv = (3/2)R (molar basis) and Cp = Cv + R = (5/2)R. Therefore, γ = Cp/Cv = (5/2)R / (3/2)R = 5/3 ≈ 1.66. Step-by-Step Solution: Cv = 3/2 * R.Cp = Cv + R = 5/2 * R.γ = Cp/Cv = (5/2 R)/(3/2 R) = 5/3 ≈ 1.66. Verification / Alternative check: Experimental values for noble gases at room temperature (e.g., argon) cluster near 1.66, validating the kinetic theory model. Why Other Options Are Wrong: 1.44 and 1.99: Not consistent with monatomic degrees of freedom. 1: Would imply Cp = Cv, which is not true for gases. Common Pitfalls: Confusing monatomic with diatomic gases (which have γ ≈ 1.4 at room temperature) and using the wrong γ in isentropic equations. Final Answer: 1.66

Question 9

Intermolecular forces from heats of vaporisation: Given ΔHvap (kJ per mol) for CS2 = 26.8, C2H5OH (ethanol) = 38.6, and H2O = 40.6, arrange the liquids in decreasing order of intermolecular forces.

Accepted Answer

Introduction / Context: The heat of vaporisation reflects the energy needed to overcome intermolecular forces from liquid to vapour. Larger ΔHvap typically indicates stronger intermolecular attractions. This principle is widely used to rationalise boiling points and solvent properties. Given Data / Assumptions: ΔHvap(H2O) = 40.6 kJ per mol, ΔHvap(EtOH) = 38.6 kJ per mol, ΔHvap(CS2) = 26.8 kJ per mol. Similar temperature reference assumed for comparison. No association changes between liquids beyond typical behaviour. Concept / Approach: Stronger intermolecular forces require more energy to vaporise a liquid. Water exhibits strong hydrogen bonding, ethanol has hydrogen bonding but weaker than water, while carbon disulfide is nonpolar with weaker dispersion forces, giving the smallest ΔHvap among the three. Step-by-Step Solution: Compare ΔHvap numerically: 40.6 (H2O) > 38.6 (EtOH) > 26.8 (CS2).Map ΔHvap magnitude to force strength.Thus, decreasing order: H2O > C2H5OH > CS2. Verification / Alternative check: Boiling points correlate: H2O boils at 100 °C, ethanol near 78 °C, CS2 near 46 °C at 1 atm—consistent with the order of intermolecular forces. Why Other Options Are Wrong: Any order placing CS2 above ethanol or water contradicts the given ΔHvap values and known polarity considerations. Common Pitfalls: Confusing polarity with hydrogen bonding strength, and ignoring that nonpolar liquids usually have lower ΔHvap unless very large molar masses are involved. Final Answer: H2O > C2H5OH > CS2

Question 10

What is the name of the unit operation that is essentially the reverse of fractional crystallisation (i.e., selective dissolution of a component from a solid)?

Accepted Answer

Introduction / Context: Fractional crystallisation separates components by crystallising one (or more) components from solution. The reverse idea is to selectively dissolve targeted constituents from a solid matrix using a suitable solvent—an operation ubiquitous in hydrometallurgy and food processing. Given Data / Assumptions: Focus on a unit operation that removes solubles from solids using a liquid solvent. Differentiation from gas–liquid mass transfer operations is required. Concept / Approach: Leaching involves contacting a solid mixture with a solvent to dissolve desired solute(s), leaving behind an insoluble residue. It is the conceptual opposite of crystallising a solute from solution to separate it in solid form. The other listed options are gas–liquid or distillation operations, not solid–liquid extraction of solubles. Step-by-Step Solution: Identify reverse of crystallisation: move from solid to solution by dissolution.Match to unit operations: leaching (solid–liquid extraction) fits this description.Exclude gas–liquid options (stripping, absorption) and distillation (vapour–liquid separation). Verification / Alternative check: Textbook definitions: Leaching (a.k.a. lixiviation) is used to recover sugar from beet, copper from ores, etc.—all involve dissolving solute from a solid matrix. Why Other Options Are Wrong: Stripping/absorption: Gas–liquid mass transfer operations. Differential distillation: Vapour–liquid separation based on volatility, not dissolution from a solid. Common Pitfalls: Confusing liquid–liquid extraction with leaching; here, the feed is a solid containing soluble species. Final Answer: leaching

Question 11

Definition check: Molality (m) is defined as the number of gram-moles of solute per what quantity of solvent?

Accepted Answer

Introduction / Context: Concentration units appear throughout reaction engineering and separation processes. Molality is particularly useful because it is independent of temperature and pressure, unlike molarity which depends on solution volume. Given Data / Assumptions: Molality symbol m. We are using standard chemical engineering definitions. Concept / Approach: Molality is defined as m = (moles of solute) / (kilograms of solvent). The denominator is the mass of the solvent only, not the total solution mass, and not the volume, which makes molality insensitive to thermal expansion or contraction. Step-by-Step Solution: State definition: m = n_solute / mass_solvent(kg).Identify the correct denominator unit: kg of solvent.Therefore, the correct choice is “kg”. Verification / Alternative check: Compare with molarity: M = moles solute per litre of solution; this varies with temperature through density changes, unlike molality. Why Other Options Are Wrong: Litre: Volume-based—this is molarity, not molality. gm mole: Dimensionally incorrect as a denominator. gm: If used, must be converted to kg; the canonical unit is kg. Common Pitfalls: Confusing solvent mass with solution mass; molality excludes the solute in the denominator. Final Answer: kg

Question 12

Psychrometrics: The saturated molal absolute humidity of a vapour–gas mixture depends on which variables?

Accepted Answer

Introduction / Context: Absolute humidity (on a molal basis) is central in humidification operations, dryer design, and environmental control. For saturated mixtures, the vapour partial pressure equals its saturation pressure, linking humidity directly to temperature and system pressure. Given Data / Assumptions: Ideal gas behaviour for both carrier gas and vapour. Mixture is saturated at the dry-bulb temperature. Total system pressure P is uniform. Concept / Approach: At saturation, y_v = p_sat(T_db) / P. The molal absolute humidity (e.g., moles vapour per mole dry gas) is a function of the saturation vapour pressure at the prevailing temperature and the total pressure. Thus both the saturation pressure and total pressure determine the saturated composition. Step-by-Step Solution: Write vapour mole fraction: y_v,sat = p_sat(T_db) / P.Relate molal humidity to y_v via standard definitions (e.g., Y = y_v / (1 − y_v)).Therefore, saturated molal absolute humidity depends on p_sat(T_db) and P. Verification / Alternative check: Psychrometric charts at different elevations (pressures) show different saturated humidity lines, confirming pressure dependence. At a fixed pressure, higher temperature increases p_sat and hence humidity. Why Other Options Are Wrong: (a) only or (b) only ignores the explicit form y_v,sat = p_sat/P, which contains both variables. Neither: Incorrect because both variables are influential. Common Pitfalls: Assuming independence from pressure (valid only at very low y_v and narrow ranges); in general, pressure matters. Final Answer: both (a) and (b)

Question 13

Diffusion-film estimate: Pure aniline evaporates through a stagnant air film of thickness 1 mm at 300 K and 100 kPa. Given p_sat(aniline) = 0.1 kPa, total molar concentration = 40.1 mol/m^3, and D_AB = 0.74 × 10^-5 m^2/s, the mass transfer coefficie…

Accepted Answer

Introduction / Context: Mass transfer coefficients can be defined with different driving forces: concentration, mole fraction, or partial pressure. Recognising the correct units for a given numerical value prevents dimensional mistakes in design equations for evaporation, drying, and gas–liquid operations. Given Data / Assumptions: Stagnant air film of thickness δ = 1 mm = 1 × 10^-3 m. Binary diffusivity D_AB = 0.74 × 10^-5 m^2/s. Reported coefficient magnitude: 7.4 × 10^-3 (value only). Film theory with concentration driving force k_c ≈ D_AB / δ applies. Concept / Approach: When the driving force is concentration difference (e.g., mol/m^3), the mass transfer coefficient k_c has units of velocity (m/s), because flux N = k_c * ΔC has units mol/m^2·s. Using the film approximation directly yields k_c = D_AB / δ with velocity units. Step-by-Step Solution: Compute k_c = D_AB / δ.k_c = (0.74 × 10^-5 m^2/s) / (1 × 10^-3 m) = 0.74 × 10^-2 m/s = 7.4 × 10^-3 m/s.Therefore, the units are m/s (a velocity). Verification / Alternative check: If a pressure driving force were used, k_G would have units mol/m^2·s·Pa. The given numeric matches D/δ, confirming a concentration-based coefficient with velocity units. Why Other Options Are Wrong: cm/s: Same dimension but wrong magnitude if units were converted without changing the number. mol/m^2·s·Pa or kmol/m^2·s·Pa: These correspond to pressure-based coefficients, which is not consistent with D/δ. Common Pitfalls: Mixing k_c, k_y, and k_G; always match the driving-force form (ΔC, Δy, or Δp) with the corresponding coefficient and its units. Final Answer: m/s

Question 14

Heat of vaporisation versus pressure: How does the latent heat of vaporisation change with increasing pressure, and what happens at the critical pressure?

Accepted Answer

Introduction / Context: Understanding how latent heat of vaporisation varies with pressure is central to phase-equilibrium calculations, boiler and condenser design, and the thermodynamics of power cycles. As a liquid approaches its critical point, the distinction between liquid and vapour fades, and this is reflected directly in the behaviour of the heat of vaporisation. Given Data / Assumptions: Latent heat of vaporisation refers to the energy required to convert saturated liquid to saturated vapour at the same temperature and pressure. System is a single-component pure fluid under equilibrium phase change conditions. Critical point is defined where the liquid and vapour phases become indistinguishable. Concept / Approach: From the Clapeyron/Clausius–Clapeyron relation, latent heat λ is tied to the slope of the saturation curve and the specific volume change between vapour and liquid. With rising pressure (and temperature along the saturation line), the difference in specific volumes shrinks. Consequently, λ decreases and must vanish at the critical point where the two phases coalesce. Step-by-Step Solution: Clausius–Clapeyron (simplified): λ = T * Δv * (dP_sat/dT).As pressure increases along saturation, Δv = v_g − v_l decreases markedly.Therefore λ decreases with pressure and tends to zero when Δv → 0 at the critical point. Verification / Alternative check: Steam tables corroborate this trend: compare λ at 1 bar to values near the critical point of water (around 22.06 MPa). The latent heat diminishes steadily and becomes zero at the critical point. Why Other Options Are Wrong: Increases: Opposite to observed data and theory. Becomes zero at critical pressure (alone): Incomplete; it also decreases continuously before reaching zero. Remains constant: Contradicted by property tables for all common fluids. Common Pitfalls: Confusing heat of vaporisation with sensible heat; assuming constant latent heat across pressures; forgetting that phase boundaries vanish at the critical point. Final Answer: both decreases and becomes zero at critical pressure

Question 15

Comparing acidity using pH: A solution with pH = 5 is less acidic than a solution with pH = 2 by what factor? (Assume 25°C and aqueous solutions.)

Accepted Answer

Introduction / Context: The pH scale is logarithmic, not linear. Every one-unit change corresponds to a tenfold change in hydrogen ion concentration. Converting qualitative statements like “more acidic” into quantitative factors is a routine task in water chemistry, bioprocessing, and environmental engineering. Given Data / Assumptions: Definition: pH = −log10[H+]. Two solutions: pH1 = 5 and pH2 = 2. Temperature and activity corrections are neglected for this comparison (typical for textbook problems). Concept / Approach: The acidity comparison factor is the ratio of hydrogen ion concentrations. Because pH is a base-10 logarithm, a difference of ΔpH corresponds to a factor of 10^ΔpH in [H+]. Lower pH means higher [H+] (more acidic). We compute the factor by evaluating 10^(pH_high − pH_low). Step-by-Step Solution: ΔpH = 5 − 2 = 3.Factor difference in [H+] = 10^3 = 1000.Therefore, the pH 5 solution is 1000 times less acidic (i.e., has 1000 times lower [H+]) than the pH 2 solution. Verification / Alternative check: Compute [H+] explicitly: at pH 2, [H+] = 10^−2 M; at pH 5, [H+] = 10^−5 M. Ratio = (10^−2)/(10^−5) = 10^3 = 1000. Why Other Options Are Wrong: 3: Mistakenly treats the scale as linear. 100: Only accounts for a 2-unit change, not the full 3 units. None of these / 30: Not supported by the logarithmic definition. Common Pitfalls: Forgetting the logarithmic nature of pH; mixing up which solution is more acidic; confusing “times less acidic” with differences in pH units. Final Answer: 1000

Question 16

Charles’ law application at constant pressure: A gas has volume 283 cm³ at 10°C. Heated to 20°C at the same pressure, what is its new volume (in cm³)?

Accepted Answer

Introduction / Context: This is a direct application of Charles’ law, which states that for a fixed amount of ideal gas at constant pressure, volume is proportional to absolute temperature. Conversions must always be done in Kelvin to avoid errors. Given Data / Assumptions: Initial volume V1 = 283 cm³. Initial temperature T1 = 10°C = 283 K. Final temperature T2 = 20°C = 293 K. Pressure is constant; gas amount unchanged; ideal gas assumption. Concept / Approach: Charles’ law: V ∝ T at constant pressure and moles. Therefore V2 = V1 * (T2 / T1), with T in Kelvin. Celsius values cannot be used directly because the proportionality holds only relative to absolute zero. Step-by-Step Solution: Convert temperatures: 10°C → 283 K; 20°C → 293 K.Apply Charles’ law: V2 = 283 cm³ * (293 / 283).Compute: V2 ≈ 293 cm³ (since the ratio is ≈ 1.0353). Verification / Alternative check: The 10 K rise near room temperature is roughly a 10/283 ≈ 3.5% increase in volume. 3.5% of 283 cm³ is ~9.9 cm³; 283 + 9.9 ≈ 292.9 cm³, agreeing with 293 cm³. Why Other Options Are Wrong: 283: Ignores temperature change. 566: Incorrectly doubles the volume (confuses proportionality). 141.5: Implies halving volume; not applicable here. 301: Overestimates the proportional increase. Common Pitfalls: Using Celsius instead of Kelvin; assuming linear changes from Celsius degrees; rounding too early. Final Answer: 293

Question 17

Percent oxygen by mass in sodium hydroxide (NaOH): Calculate the mass percent of oxygen in NaOH.

Accepted Answer

Introduction / Context: Elemental mass percent in a compound is a foundational concept for stoichiometry, analytical chemistry, and process formulation. Here we compute the percentage of oxygen in sodium hydroxide (NaOH). Given Data / Assumptions: Molar masses (approximate): Na = 23 g/mol, O = 16 g/mol, H = 1 g/mol. Compound: NaOH. Round-number atomic masses are acceptable for this calculation. Concept / Approach: Mass percent of an element = (mass of that element in one mole of the compound / molar mass of the compound) * 100%. For NaOH, we add atomic masses to get the formula mass, then compute the oxygen fraction. Step-by-Step Solution: M(NaOH) = 23 + 16 + 1 = 40 g/mol.Mass of oxygen per mole = 16 g.Percent oxygen = (16 / 40) * 100% = 40%. Verification / Alternative check: Check that the three mass percentages for Na, O, and H sum to 100%: Na = 23/40 = 57.5%; O = 40%; H = 2.5%; total = 100%. Why Other Options Are Wrong: 1, 10, 16, 25: These values do not correspond to the correct mass fraction from the formula mass breakdown. Common Pitfalls: Using atomic percent instead of mass percent; mis-adding atomic masses; forgetting to multiply by 100. Final Answer: 40

Question 18

Concentration unit definition: The number of gram-moles of solute dissolved in 1 kg of solvent is called…

Accepted Answer

Introduction / Context: Different concentration units suit different applications. Molality is particularly useful in colligative property calculations because it is independent of temperature and pressure, being based on solvent mass rather than solution volume. Given Data / Assumptions: Solute amount expressed in gram-moles (kmol or mol in SI). Reference mass of solvent is exactly 1 kg. Ambient conditions do not affect the definition since mass is invariant under thermal expansion. Concept / Approach: Molality (symbol m) is defined as moles of solute per kilogram of solvent. Contrast this with molarity (moles per litre of solution), normality (equivalents per litre), and formality (formula-gram per litre, used for ionic salts). Because solution volume can vary with temperature, molality is the preferred variable in many thermodynamic relations. Step-by-Step Clarification: Molality m = n_solute (mol) / mass_solvent (kg).Molarity M = n_solute (mol) / volume_solution (L).Normality N = equivalents / litre of solution (depends on reaction context). Verification / Alternative check: In freezing-point depression, ΔT_f = K_f * m, directly leveraging molality. This temperature independence is why tables and correlations often use molality. Why Other Options Are Wrong: Molarity/Normality/Formality: All reference solution volume, not 1 kg of solvent. Common Pitfalls: Confusing “per kg of solvent” with “per litre of solution”; using normality without defining the reaction equivalence correctly. Final Answer: molality

Question 19

Pressure unit equivalence: 1 ata (technical atmosphere, absolute) is approximately equal to which of the following?

Accepted Answer

Introduction / Context: Pressure units are numerous: atm, bar, Pa, torr, kgf/cm², and ata appear in data sheets and legacy equipment. Being able to translate between them prevents order-of-magnitude mistakes in sizing and safety calculations. Given Data / Assumptions: Standard atmosphere: 1 atm = 101.325 kPa ≈ 760 torr ≈ 1.01325 bar. Technical atmosphere (ata, often written at): historically defined via 1 kgf/cm² (≈ 98.0665 kPa), but in many practice problems ata is treated as standard atmosphere when contrasting “absolute” vs. “gauge”. Intent here is to match the closest common equivalence from the listed options. Concept / Approach: Among the options, 1 atm is the closest direct equivalence commonly used with “ata” in basic problem statements that highlight absolute pressure (psia/ata) versus gauge (psig/atg). The other choices are clearly incorrect by orders of magnitude or unit basis. Step-by-Step Reasoning: Compare magnitudes: 1 atm ≈ 101.325 kPa; 10 torr ≈ 0.013 atm (far too small).0.98 Pascal is 0.0000097 atm (off by five orders of magnitude).1 kgf/cm² ≈ 0.9678 atm (not exact; close but not equal in typical tabulations). Verification / Alternative check: Many handbooks give 1 atm = 1.033 kgf/cm²; hence 1 kgf/cm² is slightly less than 1 atm. Therefore, the best equivalence in the list is 1 atm. Why Other Options Are Wrong: 10 torr and 0.98 Pascal: Incorrect by large factors. 1 kgf/cm²: Close numerically but not equal. 1 bar: 1 bar = 100 kPa, about 0.987 atm; again not equal. Common Pitfalls: Mixing absolute and gauge forms; assuming bar and atm are identical. Final Answer: 1 atm

Question 20

Effect of changing hydrogen ion concentration on pH and acidity: A solution has pH = 5.9. If [H+] is decreased by a factor of 100, the solution becomes…

Accepted Answer

Introduction / Context: The pH scale quantifies acidity via the negative decimal logarithm of hydrogen ion concentration. Because it is logarithmic, multiplicative changes in [H+] translate to additive changes in pH. Interpreting how a change in [H+] affects where the solution lies relative to neutrality (pH 7) is a common analytical step. Given Data / Assumptions: Initial pH = 5.9. Hydrogen ion concentration is decreased by 100× (i.e., divided by 100). Aqueous solution near ambient temperature; activity effects neglected. Concept / Approach: If [H+] decreases by a factor of 10^2, pH increases by 2 units: pH_new = pH_old + 2. Values above 7 are classified as basic (alkaline), below 7 as acidic, and equal to 7 as neutral (at 25°C). Step-by-Step Solution: pH_new = 5.9 + log10(100) = 5.9 + 2 = 7.9.Since pH_new = 7.9 > 7, the solution is basic.Qualitatively, reducing [H+] shifts the balance toward hydroxide dominance, increasing pH. Verification / Alternative check: Compute [H+]: at pH 5.9, [H+] ≈ 10^−5.9 M. After dividing by 100, [H+] ≈ 10^−7.9 M, which is less than 10^−7 M (neutral at 25°C), confirming basicity. Why Other Options Are Wrong: More acidic: Would require [H+] to increase, not decrease. Neutral: Would require pH to be about 7.0, not 7.9. Same acidity: Impossible when [H+] changes by two orders of magnitude. Slightly acidic: pH 7.9 is alkaline, not acidic. Common Pitfalls: Confusing the direction of pH change with [H+] change; forgetting the logarithmic relation and using linear differences. Final Answer: basic

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