Q: Gas solubility and pressure — At a fixed temperature, how does the solubility of a gas in a liquid solvent change as pressure increases?

Introduction: Design of absorbers, carbonation processes, and oxygenation systems relies on understanding how gas solubility varies with pressure at constant temperature. Given Data / Assumptions: Temperature is held constant. Dilute gas conditions where Henry’s law applies. Solvent non-reactive with the gas. Concept / Approach: Henry’s law states that at constant temperature, the dissolved concentration of a gas in a liquid is proportional to its partial pressure in the gas phase: C = k_H * p_g (using an appropriate convention). Therefore, increasing the gas pressure increases p_g and thus increases the equilibrium solubility. Step-by-Step Solution: Identify applicable regime: dilute, non-reactive gas in liquid.Apply Henry’s law: C ∝ p_g at fixed T.Conclude that solubility increases with pressure. Verification / Alternative check: Practical examples include carbonated beverages being bottled under pressure to increase CO2 solubility. Why Other Options Are Wrong: (b), (d) Contradict Henry’s law trend. (c) Would only be true if pressure had no effect, which is not the case in this regime. (e) Henry’s constant remains central to the relationship. Common Pitfalls: Confusing Henry’s law (gas solubility) with Raoult’s law (solvent vapor pressure); forgetting temperature’s influence on k_H. Final Answer: Increases

Q: Molality calculation — A solution is prepared by dissolving 1 kilomole of a nonvolatile solute in 2000 kg of solvent. What is the molality (mol/kg) of the solution?

Introduction: Molality is a composition unit commonly used in colligative property calculations because it is independent of temperature (mass-based rather than volume-based). This problem is a direct substitution exercise. Given Data / Assumptions: Moles of solute = 1 kmol = 1000 mol. Mass of solvent = 2000 kg. Ideal dilute solution; unit consistency required. Concept / Approach: Molality m is defined as m = moles of solute / kilograms of solvent. No information about solute molar mass is needed since moles are given directly. Step-by-Step Solution: Convert 1 kmol to mol: 1 kmol = 1000 mol.Apply definition: m = 1000 mol / 2000 kg.Compute: m = 0.5 mol/kg. Verification / Alternative check: Dimensional check: (mol) / (kg) → mol/kg, correct for molality. Why Other Options Are Wrong: 2.0 and 1.0 assume fewer kilograms of solvent or more moles. 0.25 or 0.05 would require greater solvent mass (or fewer moles) than provided. Common Pitfalls: Confusing molality with molarity; forgetting to convert kmol to mol. Final Answer: 0.5

Q: Mass vs. mole fraction — An aqueous solution has a methanol mass (weight) fraction of 0.64. Which statement about the methanol mole fraction XM is correct? (Molar masses: methanol ≈ 32 g/mol, water ≈ 18 g/mol.)

Introduction: Converting between mass fraction and mole fraction is routine in chemical engineering material balances. Because different species have different molar masses, a given mass fraction rarely equals the mole fraction. Given Data / Assumptions: Methanol mass fraction w_MeOH = 0.64; water mass fraction w_H2O = 0.36. Molar masses: M_MeOH ≈ 32 g/mol, M_H2O ≈ 18 g/mol. Binary mixture only (methanol + water). Concept / Approach: Mole fraction XM = n_MeOH / (n_MeOH + n_H2O), where n_i = mass_i / M_i. Choose a convenient basis of 1 kg total solution to convert mass fractions into component masses, then into moles. Step-by-Step Solution: Take basis: 1 kg solution → m_MeOH = 0.64 kg, m_H2O = 0.36 kg.Compute moles: n_MeOH = 0.64 kg / 0.032 kg/mol = 20 mol/kg × 0.032? Simplify using grams: 0.64 kg = 640 g → n_MeOH = 640/32 = 20 mol.Compute moles: n_H2O = 0.36 kg = 360 g → n_H2O = 360/18 = 20 mol.Total moles = 20 + 20 = 40 mol → XM = 20/40 = 0.5. Verification / Alternative check: Use a fraction form directly: XM = (w_MeOH/M_MeOH) / [(w_MeOH/M_MeOH) + (w_H2O/M_H2O)] = (0.64/32)/[(0.64/32) + (0.36/18)] = 0.02/(0.02+0.02) = 0.5. Why Other Options Are Wrong: (a) and (c) contradict the exact calculation. (d) Misinterprets mass fraction as mole fraction; due to lighter water, equal moles occur at these masses. (e) 0.36 is the mass fraction of water, not the mole fraction of methanol. Common Pitfalls: Failing to use molar masses; assuming mass fraction equals mole fraction; arithmetic slips when converting units. Final Answer: XM = 0.5

Question 1

Stoichiometry of metal–acid reaction: Six grams of magnesium (atomic weight = 24) react completely with excess acid. What mass of hydrogen gas (H2) is produced (in grams)?

Accepted Answer

Introduction / Context: Metal–acid reactions liberate hydrogen gas according to the general pattern M + 2H⁺ → M²⁺ + H₂ (for a divalent metal). This problem is a straightforward stoichiometry exercise using molar masses and reaction coefficients to determine the mass of hydrogen evolved. Given Data / Assumptions: Metal: magnesium, atomic weight = 24 g/mol. Acid present in excess; magnesium is the limiting reagent. Reaction: Mg + 2H⁺ → Mg²⁺ + H₂ (1:1 molar ratio between Mg and H₂). Concept / Approach: Compute moles of Mg from its mass and molar mass. Use the stoichiometric 1:1 ratio to obtain moles of H₂. Multiply by the molar mass of H₂ (2 g/mol) to obtain mass of hydrogen produced. Step-by-Step Solution: n_Mg = 6 g / 24 g·mol^-1 = 0.25 mol.Stoichiometry: 1 mol Mg → 1 mol H₂, so n_H2 = 0.25 mol.Mass of H₂ = n_H2 * M_H2 = 0.25 mol * 2 g·mol^-1 = 0.5 g.Therefore, 0.5 g of hydrogen is produced. Verification / Alternative check: Using moles conserves atoms: magnesium atoms transfer two electrons to two protons, pairing to form one molecule of H₂ per Mg atom, confirming the 1:1 mole ratio. Why Other Options Are Wrong: 1 g or 3 g and 5 g do not match the stoichiometric calculation with correct molar masses. Common Pitfalls: Using a 2:1 ratio mistakenly (confusing with reaction coefficients). Using 1 g/mol for H rather than 2 g/mol for H₂ when converting to mass. Final Answer: 0.5

Question 2

Heat capacity of gases at pressure and critical behavior: Pick the incorrect statement regarding Cp of gases at elevated pressures and near the critical point.

Accepted Answer

Introduction / Context: Real-gas heat capacities deviate from ideal behavior as pressure increases, and they exhibit singular behavior near the critical point. Knowing which qualitative statements hold is important for process design and property estimation in compressors, heaters, and reactors. Given Data / Assumptions: Pure gases away from strong associating effects. Comparison across moderate pressures and near the critical region. Qualitative trends are sought, not exact numbers. Concept / Approach: At modest pressures and temperatures above the critical, Cp changes with pressure are usually small; many engineering calculations neglect them. Near the critical point, thermodynamic response functions such as Cp can become very large and, in the limit, diverge. A blanket claim that Cp increases substantially for all gases with pressure is incorrect and overgeneralized. Step-by-Step Solution: Assess (a): Reasonable engineering approximation—often negligible at moderate pressures above Tc.Assess (b): Overstated and not universally true; some gases show small changes or non-monotonic trends.Assess (c): Correct; Cp diverges at the critical point for pure fluids.Therefore, the single incorrect statement is (b). Verification / Alternative check: Equations of state and corresponding-states charts show Cp anomalies near criticality and mild pressure dependence away from it, validating (a) and (c) while refuting (b). Why Other Options Are Wrong: (a) and (c) are acceptable generalizations; (d) cannot be true because at least one statement ((c)) is correct. Common Pitfalls: Assuming ideal-gas Cp everywhere; ignoring critical-region anomalies. Overgeneralizing pressure effects without considering temperature relative to Tc. Final Answer: For all gases, Cp increases substantially with pressure above atmospheric pressure.

Question 3

Petroleum property correlation: API gravity (°API) is related to specific gravity G (at 15.5 °C / 60 °F) by which standard formula?

Accepted Answer

Introduction / Context: API gravity is a widely used measure of petroleum liquids, inversely related to density. Converting between specific gravity and °API is routine in refining and trading. The formula is standardized at 60 °F (15.5 °C). Given Data / Assumptions: Specific gravity G referenced to water at 60 °F (15.5 °C). Standard petroleum convention. No temperature correction needed beyond this reference. Concept / Approach: The established relationship is °API = (141.5 / G) − 131.5, yielding higher °API for lighter crudes (lower G). This transformation linearizes comparisons and is central to custody transfer specifications. Step-by-Step Solution: Recall standard: °API = (141.5 / G) − 131.5.Check limiting cases: G = 1 → °API = 10 (freshwater reference).Lighter oils (G < 1) give °API > 10, heavier give less.Select option (b). Verification / Alternative check: Industry handbooks and ASTM standards list the same expression, confirming its correctness. Why Other Options Are Wrong: Other formulas are nonstandard and produce incorrect values at the reference point. Common Pitfalls: Using density in kg/m^3 directly without converting to specific gravity at 60 °F. Applying the formula at temperatures far from 60 °F without correction factors. Final Answer: °API = (141.5 / G) − 131.5

Question 4

Heat capacity during phase equilibrium: What is the molar heat capacity of water in equilibrium with ice at constant pressure (at the melting point)?

Accepted Answer

Introduction / Context: Heat capacity at constant pressure Cp describes how much heat is required to raise the temperature of a mole of substance by 1 degree at constant pressure. At a first-order phase transition (e.g., melting), added heat changes phase at essentially constant temperature, which has implications for Cp behavior. Given Data / Assumptions: System: water in equilibrium with ice at the melting point and 1 atm. Quasi-static addition of heat. No superheating or supercooling effects. Concept / Approach: At phase equilibrium, an infinitesimal addition of heat does not increase temperature; it goes into latent heat of fusion. In the mathematical definition Cp = (dQ/dT)_p per mole, as dT approaches zero while dQ remains finite (latent heat), Cp tends to infinity. Step-by-Step Solution: At the melting point, temperature remains constant during phase change.A finite dQ = L_f is absorbed with dT ≈ 0.Therefore Cp = dQ/dT → ∞ at the transition.Select '∞' to represent divergence. Verification / Alternative check: Thermodynamic treatments of first-order transitions predict discontinuities or divergences in response functions like Cp at coexistence lines, consistent with calorimetric observations. Why Other Options Are Wrong: 0 or 1 do not reflect latent-heat absorption at constant temperature; 'None of these' is unnecessary given a correct option is available. Common Pitfalls: Confusing Cp at the phase boundary with Cp of a single phase just off the boundary. Ignoring that 'infinite' here means unbounded in the limit, not a numerically measurable infinity. Final Answer: ∞

Question 5

Psychrometrics of unsaturated mixtures: For an unsaturated vapor–gas mixture, the humid volume (mixture volume per unit mass of dry gas) increases with which parameter?

Accepted Answer

Introduction / Context: Humid volume is a key psychrometric property defined as the total volume of a vapor–gas mixture per unit mass of dry gas. It influences blower sizing, duct design, and contactor performance in drying, cooling towers, and air-conditioning systems. Given Data / Assumptions: Unsaturated mixture (below dew point of vapor). Ideal-gas approximation for mixture properties. Comparison at a fixed temperature when evaluating humidity effects. Concept / Approach: At a fixed temperature, increasing the mass of vapor (absolute humidity) per unit mass of dry gas adds more moles to the mixture, increasing total specific volume per mass of dry gas. Conversely, at fixed composition and temperature, increasing total pressure compresses the mixture, reducing specific volume (Boyle’s law behavior). Step-by-Step Solution: Humid volume v_h ≈ (R * T / P) * (1 + ω * (M_air / M_vapor)) per kg dry air (conceptually).At constant T: raising ω (absolute humidity) raises v_h.At constant T and composition: increasing P lowers v_h.Therefore, v_h increases with absolute humidity, not with total pressure. Verification / Alternative check: Psychrometric charts show lines of increasing specific volume with higher humidity ratio at a given dry-bulb temperature, validating the trend. Why Other Options Are Wrong: (a) Increasing total pressure decreases specific volume. (c) and (d) contradict the established relationships; only humidity increase raises humid volume at fixed temperature. Common Pitfalls: Mixing up effects of pressure and humidity on specific volume lines. Confusing humid volume per mass of dry gas with specific volume per mass of moist air. Final Answer: Absolute humidity at a given temperature

Question 6

Chemistry fundamentals — What are atoms of the same chemical element that have different masses called? (Note: These atoms have the same atomic number but different mass numbers due to a different number of neutrons.)

Accepted Answer

Introduction: In introductory general chemistry and nuclear chemistry, it is essential to distinguish among related terms that describe nuclides. This question targets the definition of atoms of the same element that differ in mass because of different neutron counts. Given Data / Assumptions: All atoms considered belong to the same element (same atomic number Z). Observed masses differ (different mass number A). Chemical behavior is largely similar; nuclear properties can differ. Concept / Approach: Isotopes are nuclides of the same element (same Z) but different mass numbers (A) due to varying numbers of neutrons. Because electrons govern chemistry, isotopes show nearly identical chemical properties, though physical properties (density, diffusion rates) and nuclear stability can vary. Step-by-Step Solution: Identify the defining feature: same Z (element identity), different A (mass).Recall terminology: isotopes → same Z; isobars → same A; isotones → same N (neutrons).Match the definition to the term “isotopes.” Verification / Alternative check: Examples include ¹²C, ¹³C, and ¹⁴C for carbon; ¹H, ²H (deuterium), and ³H (tritium) for hydrogen. Why Other Options Are Wrong: Isobars: same mass number but different elements (different Z). Isotones: same neutron number N but different Z. Allotropes: different structural forms of the same element in the same physical state (applies to solids like carbon, not individual atoms). “None of these” is incorrect because a standard term exists: isotopes. Common Pitfalls: Confusing isobars and isotopes; thinking allotropes refer to atomic differences rather than structural forms of elements. Final Answer: Isotopes

Question 7

Colligative properties — If you prepare separate solutions by dissolving equimolar amounts of different non-electrolyte solutes in the same mass of a given solvent, what will be the elevation in boiling point for each solution (at the same pressure)?

Accepted Answer

Introduction: Boiling-point elevation is a classic colligative property. Colligative properties depend only on the number of solute particles present, not their chemical identity (assuming ideal dilute solutions and no association/dissociation). Given Data / Assumptions: Equimolar amounts (same moles) of different solutes are dissolved. The same mass of the same solvent is used for each solution. Solutes are non-electrolytes (van’t Hoff factor i ≈ 1) and solutions are dilute. Pressure is the same for all cases. Concept / Approach: Boiling-point elevation is given by ΔT_b = i * K_b * m, where m is molality and K_b is the ebullioscopic constant of the solvent. With equal moles of solute in the same solvent mass, m is identical for all solutions. With i ≈ 1 and the same K_b (same solvent), ΔT_b is the same for each solution, regardless of solute identity. Step-by-Step Solution: Set m = n_solute / kg_solvent; with same n_solute and same kg_solvent → same m for all.Use ΔT_b = i * K_b * m; with i ≈ 1 and same K_b → equal ΔT_b.Conclude that each solution exhibits the same boiling-point elevation. Verification / Alternative check: If a solute dissociates (i > 1) or associates (i < 1), ΔT_b would differ. The problem context specifies different non-electrolytes to keep i ≈ 1. Why Other Options Are Wrong: (b) Conflicts with the definition of a colligative property under the stated assumptions. (c) ΔT_b does not scale with molecular weight; it scales with particle count (molality). (d) Boiling point does elevate for nonvolatile solutes. (e) K_b is a constant for the solvent and is the same in each case, not a variable linking differences here. Common Pitfalls: Forgetting van’t Hoff factors for electrolytes; mixing up molarity and molality in boiling-point elevation problems. Final Answer: The same elevation in boiling point for all solutions

Question 8

Gas solubility and pressure — At a fixed temperature, how does the solubility of a gas in a liquid solvent change as pressure increases?

Accepted Answer

Introduction: Design of absorbers, carbonation processes, and oxygenation systems relies on understanding how gas solubility varies with pressure at constant temperature. Given Data / Assumptions: Temperature is held constant. Dilute gas conditions where Henry’s law applies. Solvent non-reactive with the gas. Concept / Approach: Henry’s law states that at constant temperature, the dissolved concentration of a gas in a liquid is proportional to its partial pressure in the gas phase: C = k_H * p_g (using an appropriate convention). Therefore, increasing the gas pressure increases p_g and thus increases the equilibrium solubility. Step-by-Step Solution: Identify applicable regime: dilute, non-reactive gas in liquid.Apply Henry’s law: C ∝ p_g at fixed T.Conclude that solubility increases with pressure. Verification / Alternative check: Practical examples include carbonated beverages being bottled under pressure to increase CO2 solubility. Why Other Options Are Wrong: (b), (d) Contradict Henry’s law trend. (c) Would only be true if pressure had no effect, which is not the case in this regime. (e) Henry’s constant remains central to the relationship. Common Pitfalls: Confusing Henry’s law (gas solubility) with Raoult’s law (solvent vapor pressure); forgetting temperature’s influence on k_H. Final Answer: Increases

Question 9

Molality calculation — A solution is prepared by dissolving 1 kilomole of a nonvolatile solute in 2000 kg of solvent. What is the molality (mol/kg) of the solution?

Accepted Answer

Introduction: Molality is a composition unit commonly used in colligative property calculations because it is independent of temperature (mass-based rather than volume-based). This problem is a direct substitution exercise. Given Data / Assumptions: Moles of solute = 1 kmol = 1000 mol. Mass of solvent = 2000 kg. Ideal dilute solution; unit consistency required. Concept / Approach: Molality m is defined as m = moles of solute / kilograms of solvent. No information about solute molar mass is needed since moles are given directly. Step-by-Step Solution: Convert 1 kmol to mol: 1 kmol = 1000 mol.Apply definition: m = 1000 mol / 2000 kg.Compute: m = 0.5 mol/kg. Verification / Alternative check: Dimensional check: (mol) / (kg) → mol/kg, correct for molality. Why Other Options Are Wrong: 2.0 and 1.0 assume fewer kilograms of solvent or more moles. 0.25 or 0.05 would require greater solvent mass (or fewer moles) than provided. Common Pitfalls: Confusing molality with molarity; forgetting to convert kmol to mol. Final Answer: 0.5

Question 10

Mass vs. mole fraction — An aqueous solution has a methanol mass (weight) fraction of 0.64. Which statement about the methanol mole fraction XM is correct? (Molar masses: methanol ≈ 32 g/mol, water ≈ 18 g/mol.)

Accepted Answer

Introduction: Converting between mass fraction and mole fraction is routine in chemical engineering material balances. Because different species have different molar masses, a given mass fraction rarely equals the mole fraction. Given Data / Assumptions: Methanol mass fraction w_MeOH = 0.64; water mass fraction w_H2O = 0.36. Molar masses: M_MeOH ≈ 32 g/mol, M_H2O ≈ 18 g/mol. Binary mixture only (methanol + water). Concept / Approach: Mole fraction XM = n_MeOH / (n_MeOH + n_H2O), where n_i = mass_i / M_i. Choose a convenient basis of 1 kg total solution to convert mass fractions into component masses, then into moles. Step-by-Step Solution: Take basis: 1 kg solution → m_MeOH = 0.64 kg, m_H2O = 0.36 kg.Compute moles: n_MeOH = 0.64 kg / 0.032 kg/mol = 20 mol/kg × 0.032? Simplify using grams: 0.64 kg = 640 g → n_MeOH = 640/32 = 20 mol.Compute moles: n_H2O = 0.36 kg = 360 g → n_H2O = 360/18 = 20 mol.Total moles = 20 + 20 = 40 mol → XM = 20/40 = 0.5. Verification / Alternative check: Use a fraction form directly: XM = (w_MeOH/M_MeOH) / [(w_MeOH/M_MeOH) + (w_H2O/M_H2O)] = (0.64/32)/[(0.64/32) + (0.36/18)] = 0.02/(0.02+0.02) = 0.5. Why Other Options Are Wrong: (a) and (c) contradict the exact calculation. (d) Misinterprets mass fraction as mole fraction; due to lighter water, equal moles occur at these masses. (e) 0.36 is the mass fraction of water, not the mole fraction of methanol. Common Pitfalls: Failing to use molar masses; assuming mass fraction equals mole fraction; arithmetic slips when converting units. Final Answer: XM = 0.5

Question 11

Real vs. ideal gases — A truly ideal gas cannot be liquefied. The primary reason is that, in the ideal-gas model,

Accepted Answer

Introduction: Liquefaction of gases depends on intermolecular attractions strong enough to hold molecules together as a condensed phase. This question contrasts ideal-gas assumptions with the behavior required for liquefaction. Given Data / Assumptions: Ideal gas model neglects molecular size and attractions (no a, b terms as in van der Waals). Liquefaction requires a finite critical temperature and pressure determined by real interactions. Concept / Approach: In the ideal-gas limit, attractive forces are zero. Without attractions, there is no driving force for condensation, so an ideal gas cannot exhibit a liquid phase or a finite critical point. Real gases deviate from ideality precisely because of these interactions, enabling liquefaction below the critical temperature. Step-by-Step Solution: Recall ideal-gas postulates: no intermolecular forces; point-like molecules.Relate liquefaction to attractions that create a potential well facilitating condensation.Conclude the key reason: absence of intermolecular attractions in the ideal model. Verification / Alternative check: The van der Waals equation incorporates parameters a (attractions) and b (co-volume). Setting a = 0 removes the possibility of a liquid phase. Why Other Options Are Wrong: (a) Small size alone does not prevent liquefaction; noble gases are small yet liquefy. (b) Critical temperature varies by gas and is not linked to 0°C specifically. (d) Gases do not “skip” the liquid phase under normal conditions; sublimation pertains to solids. (e) Molecular rotation is not the governing factor for liquefaction. Common Pitfalls: Assuming liquefaction depends on absolute temperature markers like 0°C; conflating idealized models with real behavior. Final Answer: intermolecular forces are negligible (no attractions)

Question 12

Water quality — A well-water sample contains 140 g/m³ of Ca²⁺ ions and 345 g/m³ of Na⁺ ions. What is the hardness of this water expressed as equivalent CaCO₃ (g/m³)? (Atomic masses: Ca = 40, Na = 23, C = 12, O = 16.)

Accepted Answer

Introduction: Hardness expresses the concentration of multivalent cations (mainly Ca²⁺ and Mg²⁺) as an equivalent amount of CaCO3. It is widely used in water-treatment design and reporting. Given Data / Assumptions: [Ca²⁺] = 140 g/m³ = 140 mg/L. [Na⁺] = 345 g/m³ = 345 mg/L (does not contribute to hardness). Equivalent weight: Ca²⁺ → 40/2 = 20 g/eq; CaCO3 → 50 g/eq. Concept / Approach: Hardness as CaCO3 is computed by converting the Ca²⁺ concentration to equivalents, then to CaCO3 using 50 g/eq. Sodium does not contribute to hardness because it is monovalent. Step-by-Step Solution: Convert Ca²⁺ to mg/L: 140 g/m³ = 140 mg/L.Compute hardness as CaCO3: Hardness = (mg/L of Ca²⁺) * (50 / equivalent weight of Ca²⁺).Equivalent weight of Ca²⁺ = 40/2 = 20 g/eq.Hardness = 140 * (50/20) = 140 * 2.5 = 350 mg/L.In g/m³, the numeric value is the same: 350 g/m³. Verification / Alternative check: Dimensional consistency: mg/L × (dimensionless 50/20) gives mg/L as CaCO3; converting to g/m³ preserves the number. Why Other Options Are Wrong: 485 assumes Na⁺ contributes like Ca²⁺; it does not. 140 uses Ca²⁺ as if already CaCO3 equivalents (missing the 50/20 factor). 345 incorrectly uses Na⁺ concentration as hardness. 700 would double-count or misapply equivalent weights. Common Pitfalls: Including Na⁺/K⁺ in hardness; using molar instead of equivalent weights; unit confusion between mg/L and g/m³. Final Answer: 350

Question 13

Identify the incorrect statement — Choose the one option below that is NOT correct.

Accepted Answer

Introduction: This mixed-concepts question probes knowledge of specific heats, Avogadro’s law, molar volume at standard conditions, and the definition of an azeotrope. Only one statement is intentionally incorrect. Given Data / Assumptions: NTP molar volume ≈ 22.4 L per mole of an ideal gas. Hydrogen is diatomic (H₂) under standard conditions. Azeotrope: vapor and liquid have the same composition at the boiling point for a given pressure. Concept / Approach: At NTP, 22.4 L of any ideal gas contains 1 mole of molecules, i.e., 6.023 × 10^23 molecules. For H₂, that corresponds to 2 × 6.023 × 10^23 atoms. Therefore, saying 22.4 L of H₂ contains 6.023 × 10^23 atoms is incorrect; it contains that number of molecules, not atoms. Step-by-Step Solution: Use Avogadro’s law: equal volumes at same T, P have equal number of molecules.Compute for hydrogen: 22.4 L → 1 mol molecules → 6.023 × 10^23 molecules.Atoms present = 2 per molecule → 1.2046 × 10^24 atoms, not 6.023 × 10^23. Verification / Alternative check: Check standard molar volume tables: 22.414 L at 0°C and 1 atm equals 1 mol of gas; terminology distinguishes molecules vs. atoms for diatomic gases. Why Other Options Are Wrong or Right: (a) True: extra rotational/vibrational degrees raise heat capacity for diatomic gases. (b) True by Avogadro’s law when T and P are equal. (d) True definition of an azeotrope. (e) False because one incorrect statement exists, namely (c). Common Pitfalls: Confusing “molecules” and “atoms” for diatomic gases; minor variations of standard molar volume do not change the conclusion. Final Answer: A volume of 22.4 L of hydrogen gas (H₂) at NTP contains 6.023 × 10^23 hydrogen atoms

Question 14

Ideal-gas density — Assuming CO₂ behaves as a perfect gas, compute its density (kg/m³) at 263°C and 2 atm.

Accepted Answer

Introduction: Estimating gas density from state variables using the ideal-gas relation is a frequent step in sizing equipment and calculating mass flow rates. Here we evaluate CO₂ at elevated temperature and moderate pressure. Given Data / Assumptions: T = 263°C = 263 + 273 = 536 K. P = 2 atm = 2 × 101325 Pa = 202650 Pa. Molar mass of CO₂ = 44 g/mol = 0.044 kg/mol. Ideal-gas behavior (Z = 1). Concept / Approach: Use ρ = (P * M) / (R * T), where R = 8.314 kJ/(kmol·K) expressed as 8.314 J/(mol·K) for SI consistency → 8.314 (Pa·m³)/(mol·K). Step-by-Step Solution: Compute numerator: P * M = 202650 Pa * 0.044 kg/mol = 8916.6 Pa·kg/mol.Compute denominator: R * T = 8.314 * 536 = 4459.9 Pa·m³/(mol).Compute density: ρ = 8916.6 / 4459.9 ≈ 2.00 kg/m³. Verification / Alternative check: A quick check with ρ ≈ (P/RT) * M: at 1 atm and 300 K, CO₂ is about 1.84 kg/m³; doubling pressure and raising T to 536 K yields approximately (2/ (536/300)) × 1.84 ≈ (2 * 300 / 536) × 1.84 ≈ 2.06 kg/m³ → consistent with 2. Why Other Options Are Wrong: 1 or 3 or 4 or 5 kg/m³ do not satisfy the ideal-gas calculation with given T and P. Common Pitfalls: Forgetting to convert Celsius to Kelvin; mixing units (atm vs. Pa); using molar mass in g/mol without converting to kg/mol. Final Answer: 2

Question 15

Atmospheric composition — What is the average molecular weight (apparent molar mass) of dry air typically used in engineering calculations?

Accepted Answer

Introduction: The average molecular weight of dry air is required in numerous calculations, including gas densities, flow metering, and psychrometrics. Given Data / Assumptions: Dry air composition approximated as ~78% N₂ (28 g/mol), ~21% O₂ (32 g/mol), with small fractions of Ar (40 g/mol) and CO₂ (44 g/mol). Humidity neglected (dry basis). Concept / Approach: The apparent molecular weight is a mole-fraction-weighted average of component molar masses. Using standard atmospheric proportions yields a value close to 28.96 g/mol, commonly rounded to 29 g/mol for engineering work. Step-by-Step Solution: Approximate calculation: M_air ≈ 0.7828 + 0.2132 + 0.01*40 ≈ 21.84 + 6.72 + 0.40 ≈ 28.96 g/mol.Round to the standard engineering value: 29 g/mol.Use 29 g/mol in density and molar flow computations unless high precision with real composition is needed. Verification / Alternative check: Textbook tables and property databases list dry air ≈ 28.96 g/mol (often rounded to 29 g/mol). Why Other Options Are Wrong: 21 and 23 are too low; 79 confuses with nitrogen volume percent; 32 corresponds to oxygen’s molar mass. Common Pitfalls: Using 28 g/mol (nitrogen) instead of the mixture average; neglecting argon and CO₂ contributions when high accuracy is required. Final Answer: 29

Question 16

Aqueous equilibria — At room temperature, the ionic product of water is [H+][OH−] = 10^-14 (mol/L)^2. If a solution has [OH−] = 10^-6 mol/L, what will be the pH of the solution?

Accepted Answer

Introduction / Context: pH and pOH are cornerstone ideas in aqueous chemistry and environmental engineering. At a given temperature (commonly 25°C, i.e., room temperature), the self-ionization of water imposes a fixed product [H+][OH−] known as the ionic product (Kw). This question checks your fluency in moving between hydroxide concentration, hydrogen ion concentration, and pH using Kw and logarithms, which is essential for process calculations in water treatment, biochemical reactions, and acid–base titrations. Given Data / Assumptions: At room temperature: Kw = [H+][OH−] = 1.010^-14 (mol/L)^2. The solution has [OH−] = 1.010^-6 mol/L. Activities are approximated by concentrations (dilute solution assumption). Concept / Approach: The relationships are: pH = −log10[H+], pOH = −log10[OH−], and pH + pOH = 14 (at 25°C). Given [OH−], we can compute either [H+] via Kw or directly compute pOH and then pH. Both routes should agree. Ensure unit consistency (mol/L) and base-10 logarithms. Step-by-Step Solution: Compute [H+] from Kw: [H+] = Kw / [OH−] = (1.010^-14) / (1.010^-6) = 1.010^-8 mol/L.Compute pH: pH = −log10(1.010^-8) = 8.00.Alternative method: pOH = −log10(1.0*10^-6) = 6.00; therefore pH = 14 − 6 = 8.00. Verification / Alternative check: Both methods yield the same pH = 8.00, confirming internal consistency. This also matches the expectation that [OH−] above 10^-7 mol/L leads to a basic solution (pH > 7) at 25°C. Why Other Options Are Wrong: 6: Would imply [H+] = 10^-6 mol/L, which contradicts the given [OH−]. 10 or 12: These correspond to much larger [OH−] (10^-4 or 10^-2 mol/L). The “Not defined” choice is incorrect since Kw at room temperature is well-established. Common Pitfalls: Forgetting that pH + pOH = 14 only strictly holds at 25°C; using natural logs instead of base-10; confusing units; or assuming pH 7 is always neutral regardless of temperature (Kw varies with temperature). Final Answer: 8

Question 17

Acid–base scale — If the pH of an acidic solution is decreased from 5 to 2, by what factor does the hydrogen ion concentration [H+] increase?

Accepted Answer

Introduction / Context: The pH scale is logarithmic, not linear. Every 1-unit change in pH corresponds to a tenfold change in hydrogen ion concentration [H+]. Understanding this is vital for analyzing acid strength, dosing neutralizing agents, and interpreting biological pH control. This question tests your ability to translate a pH change into the corresponding multiplicative change in [H+]. Given Data / Assumptions: Initial pH = 5; final pH = 2. Definition: pH = −log10[H+]. Temperature is constant (so that the pH scale interpretation is consistent). Concept / Approach: Because pH is the negative base-10 logarithm of [H+], a drop in pH by ΔpH units multiplies [H+] by 10^(ΔpH). Here, the decrease from 5 to 2 is 3 units, so [H+] increases by 10^3. This exponential nature explains why small pH shifts can have large chemical or biological effects. Step-by-Step Solution: Let [H+]_1 correspond to pH 5: [H+]_1 = 10^-5 mol/L.Let [H+]_2 correspond to pH 2: [H+]_2 = 10^-2 mol/L.Compute the ratio: [H+]_2 / [H+]_1 = (10^-2)/(10^-5) = 10^3 = 1000. Verification / Alternative check: Directly from the definition: change in pH = −log10([H+]_2) + log10([H+]_1) = log10([H+]_1/[H+]_2). Rearranging gives [H+]_2/[H+]_1 = 10^(pH_1 − pH_2) = 10^(5 − 2) = 10^3. Why Other Options Are Wrong: 10 or 100: These correspond to 1- or 2-unit changes, respectively. 10000 would require a 4-unit change. “Unchanged” is incorrect because pH changed significantly. Common Pitfalls: Treating pH changes linearly; forgetting that lower pH means higher [H+]; mixing up pH and pOH relationships (pH + pOH = 14 at 25°C). Final Answer: 1000

Question 18

Concentration measures — On heating an aqueous solution, which concentration unit decreases due to thermal expansion of the solution volume?

Accepted Answer

Introduction / Context: Chemical engineers frequently choose among several concentration units. Some depend on mass, others depend on volume. Since volume changes with temperature (thermal expansion or contraction), volume-based measures respond to temperature changes while mass-based measures do not. This question identifies which common unit decreases upon heating an aqueous solution. Given Data / Assumptions: Aqueous solution heated (temperature increases). Solute amount and solvent amount are fixed (no evaporation or addition). Solution expands upon heating (volume increases). Concept / Approach: Molarity (mol/L) depends on the solution volume in the denominator. When temperature rises, solution volume typically increases due to thermal expansion, so molarity falls if the amount of solute is fixed. In contrast, molality (mol/kg solvent), mole fraction (moles basis), and weight percent (mass basis) do not involve solution volume; therefore, they remain essentially unchanged with temperature (ignoring tiny mass changes). Step-by-Step Solution: Define molarity: M = moles of solute / liters of solution.As T ↑, solution volume ↑ → denominator ↑ → M ↓.Molality m = moles solute / kg solvent → mass-based, insensitive to T.Mole fraction and weight percent likewise are mass/mole based → temperature independent. Verification / Alternative check: Laboratory practice: volumetric flasks are calibrated at a reference temperature (often 20°C). Accurate molarity preparations require temperature control; molality and weight percent do not. Why Other Options Are Wrong: A/B/C/E: These measures do not include the solution volume explicitly and hence do not systematically decrease when temperature rises. Common Pitfalls: Using molarity in reactions with significant temperature change without correcting for density/volume; confusing molarity with molality due to similar names. Final Answer: molarity

Question 19

Atomic structure — The chemical nature of an element (its characteristic chemistry) is independent of which subatomic property?

Accepted Answer

Introduction / Context: An element’s chemical behavior is determined primarily by its electrons, which in turn are governed by the nuclear charge (atomic number, Z). Isotopes of the same element share the same atomic number but differ in neutron count. This question probes your understanding that isotopic variation rarely alters core chemical properties, even though it can influence physical properties and reaction rates slightly (isotope effects). Given Data / Assumptions: Chemical nature refers to typical valence, bonding, and reactivity patterns. Atomic number Z defines the number of protons and the electron structure in neutral atoms. Neutrons affect mass number and nuclear stability but not the valence shell structure directly. Concept / Approach: Since chemical reactions involve electron exchange and orbital interactions, the primary determinant is the electron configuration fixed by Z. Changing neutrons does not change Z; therefore, isotopes exhibit nearly identical chemistry. Notable exceptions include kinetic isotope effects (e.g., protium vs deuterium) which change reaction rates but typically not the kinds of compounds formed. Step-by-Step Solution: Recognize that atomic number (protons) sets electron count in neutral atoms.Electron configuration defines valence and chemical bonding.Neutron number varies (isotopes) without altering valence shell occupation.Therefore chemical nature is independent of neutron count. Verification / Alternative check: Compare isotopes of chlorine (Cl-35 and Cl-37): identical chemistry, slightly different densities/masses. Heavy water (D2O) shows rate differences and some physical differences, yet oxygen chemistry remains that of oxygen. Why Other Options Are Wrong: A/B/E: These directly affect electron structure and hence chemical behavior. D is false because there is indeed one property (neutron count) that does not determine chemical nature in the usual sense. Common Pitfalls: Confusing nuclear stability (radioactivity) with chemical reactivity; assuming any subatomic change alters chemical nature drastically—chemistry is dominantly about electrons. Final Answer: the number of neutrons present in it

Question 20

Thermochemistry — Which of the following processes is exothermic under standard conditions?

Accepted Answer

Introduction / Context: Determining whether a reaction is exothermic or endothermic is central to reactor design, safety, and energy balances. Exothermic reactions release heat (negative enthalpy change), while endothermic reactions absorb heat. This question asks you to evaluate several well-known transformations and identify if any are exothermic as stated. Given Data / Assumptions: Standard thermochemical data trends apply. No catalysts or alternate pathways that could change overall enthalpy sign. Physical states are standard (graphite is the stable allotrope of carbon at STP). Concept / Approach: Graphite to diamond raises enthalpy (diamond is metastable and higher in energy) → endothermic. Splitting water (electrolysis) requires energy input to break strong O–H bonds → endothermic. Dehydrogenation removes H2 from alkanes to form alkenes; this generally absorbs heat (breaking C–H and forming a C=C and H–H results in positive enthalpy) → endothermic. Hence, none of the listed processes are exothermic as written. Step-by-Step Solution: Graphite → diamond: ΔH > 0 (diamond less stable than graphite at 1 atm).H2O(l) → H2(g) + 1/2 O2(g): ΔH > 0 (bond breaking dominates).C2H6 → C2H4 + H2: ΔH > 0 (typical dehydrogenations are endothermic).Therefore, no process listed is exothermic → choose “none of these”. Verification / Alternative check: Standard enthalpies of formation confirm positive ΔH for these transformations. Industrial dehydrogenations are run in fired heaters to supply heat. Why Other Options Are Wrong: A/B/C are individually endothermic. Option E (not among the original choices) can be exothermic but is not offered as the principal selection; the correct answer among the listed is “none of these”. Common Pitfalls: Assuming all phase or structural changes are exothermic; overlooking that bond-breaking steps generally require input energy. Final Answer: none of these

Stoichiometry Questions