Q: Corresponding states — Gases at the same reduced temperature (Tr) and reduced pressure (Pr) will exhibit which common behavior?

Introduction / Context:The principle of corresponding states underpins generalized compressibility charts and equations of state. By normalizing temperature and pressure with respect to critical properties, many gases exhibit similar non-ideal behavior at the same reduced conditions. This allows engineers to estimate deviations from ideality without species-specific data for every case. Given Data / Assumptions: Reduced temperature Tr = T/Tc and reduced pressure Pr = P/Pc. Gases are not in reactive mixtures; behavior is compared at identical Tr and Pr. Acentric factor effects are modest (first-order corresponding states). Concept / Approach: At the same Tr and Pr, many gases show nearly the same compressibility factor Z. Similar Z implies a similar degree of deviation from ideal-gas behavior because Z quantifies PV/(nR*T). While exact equality is not guaranteed (acentric factor ω refines predictions), the “nearly same” behavior forms the basis of generalized charts and correlations. Step-by-Step Solution: Normalize conditions: compute Tr, Pr for each gas.Use generalized Z charts: at the same Tr, Pr, read Z values which cluster for many gases.Infer deviation from ideal gas: deviation ∝ |Z − 1| is similar among gases at same reduced conditions.Hence statements (a) and (b) are both valid. Verification / Alternative check: Comparisons across nitrogen, methane, and argon at identical Tr and Pr show comparable Z values, validating the principle, with small adjustments captured by ω in the Lee–Kesler method. Why Other Options Are Wrong: D rejects a well-founded principle. E concerns critical constants themselves, not reduced-condition behavior. Common Pitfalls: Using corresponding states near phase boundaries without caution; neglecting acentric factor corrections for polar or associating molecules. Final Answer: both (a) and (b)

Q: Non-ideality in liquids — The activity coefficient γ of a component in a liquid solution accounts for departure from ideal solution behavior. Which statement is true?

Introduction / Context:Real liquid mixtures deviate from Raoult’s-law ideality due to molecular size, shape, and intermolecular forces. The activity coefficient γ modifies the effective concentration (activity) to capture these deviations. Understanding what γ depends on, and what it does not directly represent, is essential for vapor–liquid equilibrium (VLE) and separation process design. Given Data / Assumptions: Activity a_i = γ_i * x_i for liquid-phase component i (at moderate pressures). γ_i depends on composition and temperature (and sometimes pressure weakly). Colligative properties are separate phenomena emergent from solution behavior, not direct definitions of γ. Concept / Approach: In ideal solutions, γ_i = 1. In non-ideal solutions, γ_i ≠ 1 and varies with the mixture composition and temperature because local molecular environments change with composition and thermal energy. While boiling point elevation or freezing point depression can be predicted using activities, γ itself is not “the measure” of those properties; rather, such properties emerge from the full thermodynamic relations. Step-by-Step Solution: Define activity: a_i = γ_i * x_i.Note dependencies: γ_i = f(T, x⃗ , sometimes P).Relate to VLE: y_i * P = x_i * γ_i * P_i^sat (modified Raoult’s law).Conclude: γ_i fundamentally varies with composition (and temperature), not a fixed constant. Verification / Alternative check: Models such as Wilson, NRTL, UNIQUAC compute γ_i as functions of composition and temperature, fitting binary parameters to data. Why Other Options Are Wrong: A/D: Boiling point elevation and freezing point depression are colligative outcomes, not definitions of γ. B: γ typically changes with T. E: γ = 1 only for ideal behavior; many mixtures are non-ideal. Common Pitfalls: Using constant γ over wide composition ranges; ignoring temperature dependence in design calculations. Final Answer: It is a function of the liquid-phase composition

Q: Gas mixtures — Equal masses of methane (CH4) and hydrogen (H2) are mixed in an empty container. What fraction of the total pressure is contributed by hydrogen (its partial pressure/total pressure)?

Introduction / Context:Dalton’s law of partial pressures states that, for ideal gases, each species’ partial pressure is proportional to its mole fraction in the mixture at a given temperature and total pressure. When mixing equal masses of gases with different molar masses, the lighter gas will contribute more moles—and hence a larger share of the total pressure. This question quantifies that share for H2 and CH4. Given Data / Assumptions: Equal masses of CH4 and H2 are mixed. Ideal-gas behavior; same temperature and total pressure. Molar masses: M_CH4 = 16 g/mol; M_H2 = 2 g/mol. Concept / Approach: Partial pressure fraction = mole fraction for ideal gases. With equal masses, moles are inversely proportional to molar mass. Because hydrogen is eight times lighter than methane, equal masses contain eight times more moles of H2 than CH4, so H2 dominates the mole fraction and therefore the partial pressure. Step-by-Step Solution: Let each gas mass be m grams.Moles of CH4 = m/16; moles of H2 = m/2 = 8*(m/16).Total moles = (m/16) + (8m/16) = 9m/16.Mole fraction of H2 = (8m/16) / (9m/16) = 8/9.Therefore partial pressure of H2 = (8/9)*P_total. Verification / Alternative check: Dalton’s law: p_i = y_i * P_total. The computed mole fraction 8/9 directly sets the partial pressure fraction. Why Other Options Are Wrong: 1/2 suggests equal moles (not the case). 1/9 would correspond to methane’s share. 5/9 or 7/9 are arbitrary and not supported by stoichiometry. Common Pitfalls: Confusing equal masses with equal moles; ignoring the large molar-mass difference which skews mole fractions heavily toward the lighter gas. Final Answer: 8/9

Q: Heat capacity integration — Air has Cp ≈ 26.693 + 7.365×10^-3·T (J/mol·K), with T in K. How much heat (kJ) does 1 mol of air give off when cooled at constant pressure from 500°C to −100°C?

Introduction / Context:When specific heat depends on temperature, accurate enthalpy changes require integrating Cp over the temperature range. This is common in combustion products, gas cooling, and cryogenic operations. Here, you must perform a straightforward definite integral for Cp(T) over the cooling path to find the heat released by 1 mol of air at constant pressure. Given Data / Assumptions: Cp(T) = 26.693 + 7.36510^-3 * T (J/mol·K). Initial temperature T1 = 500°C = 773.15 K. Final temperature T2 = −100°C = 173.15 K. Constant pressure process; ideal-gas approximation for air. Concept / Approach: For a constant-pressure process, the molar enthalpy change is ΔH = ∫(from T2 to T1) Cp(T) dT if you want the magnitude of heat released (positive). The sign convention for cooling is that the system releases heat: q_p = −∫(from T2 to T1) Cp dT if proceeding from hot to cold. We will compute the magnitude of heat given off. Step-by-Step Solution: Compute ΔT = T1 − T2 = 773.15 − 173.15 = 600.00 K.Integrate Cp(T): ∫Cp dT = ∫[26.693 + 7.36510^-3T] dT = 26.693(ΔT) + (7.36510^-3/2)(T1^2 − T2^2).Calculate the two terms: 26.693600 = 16015.8 J/mol; (7.36510^-3/2)*(773.15^2 − 173.15^2) ≈ 2090.85 J/mol.Sum: 16015.8 + 2090.85 ≈ 18106.65 J/mol = 18.10665 kJ/mol.Rounded to two decimal places: 18.11 kJ/mol of heat released. Verification / Alternative check: A linear Cp(T) gives a quadratic antiderivative. A quick midpoint estimate using Cp at ~473 K also gives a close value, reinforcing the integral result. Why Other Options Are Wrong: 10.73 and 16.15 kJ underestimate the integrated Cp; 18.33 kJ slightly overestimates; 20.50 kJ is too high for the specified range and Cp. Common Pitfalls: Using constant Cp instead of integrating; forgetting unit conversions from J to kJ; mixing Celsius with Kelvin in Cp expressions. Final Answer: 18.11 kJ

Q: Acid–base calculations – relation between pH and pOH For an aqueous solution at 25°C with pH = 8, determine the corresponding pOH value using the water ion-product relationship.

Introduction / Context:pH and pOH are complementary measures of acidity and basicity in water. At 25°C, their sum is fixed by the ionic product of water. This quick relation is used constantly in environmental engineering, bioprocessing, and analytical chemistry. Given Data / Assumptions: Temperature 25°C (298 K). pH = 8 given. Pure water reference where Kw = 1.0 * 10^-14 at 25°C. Concept / Approach:By definition, pH = -log10[H+] and pOH = -log10[OH-]. The ion product is Kw = [H+]*[OH-] = 1.0 * 10^-14 at 25°C, which leads to pH + pOH = 14. Hence knowing one immediately gives the other by subtraction. Step-by-Step Solution: Use identity: pH + pOH = 14 at 25°C.Substitute pH = 8 → pOH = 14 - 8.Compute pOH = 6. Verification / Alternative check:[H+] = 10^-8 M → [OH-] = Kw / [H+] = 10^-14 / 10^-8 = 10^-6 M → pOH = 6. Consistent. Why Other Options Are Wrong: 1, 7, 10: Do not satisfy pH + pOH = 14 for pH = 8. Common Pitfalls:Using pH + pOH = 14 at temperatures far from 25°C without adjusting Kw; being careless with base-10 logarithms. Final Answer:6

Q: Osmotic pressure of dilute solutions – proportionality for a Raoult's-law system For a dilute solution of a non-volatile solute obeying Raoult's law, the osmotic pressure is directly proportional to which single parameter (holding the others constan…

Introduction / Context:Osmotic pressure is a colligative property used to estimate molecular weights and design membrane separations. For dilute, ideal solutions, a simple gas-law-like expression connects osmotic pressure to solute amount, solution volume, and absolute temperature. Given Data / Assumptions: Dilute solution of non-volatile solute. Raoult's-law (ideal) behavior is valid. Van't Hoff equation applies. Concept / Approach:The van't Hoff relation is πV = nRT, or π = (n/V) * R * T. Thus π is directly proportional to temperature T at fixed solute amount per unit volume, directly proportional to solute moles n at fixed V and T, and inversely proportional to volume V at fixed n and T. Because the question requests a single proportional factor, the most universal statement among the options is direct proportionality to absolute temperature T. Step-by-Step Solution: Write the relation: π = (n/V) * R * T.Hold n/V constant to isolate the proportional dependence on T.Conclude π ∝ T under the stated ideal, dilute conditions. Verification / Alternative check:Experimentally, increasing temperature at constant concentration raises measured osmotic pressure in line with the gas constant scaling. Why Other Options Are Wrong: Volume of solution: π is inversely proportional to V at fixed n and T, not directly proportional. Moles of solute: True only if V and T are held constant; the prompt asks for a single proportionality, and temperature is the standard statement. None of these: Incorrect because π does scale directly with T for ideal solutions. Common Pitfalls:Failing to state the ceteris paribus condition; forgetting that osmotic pressure depends on concentration and temperature simultaneously. Final Answer:temperature.

Q: Water properties – saturated vapor pressure at the normal boiling point At 100°C (the normal boiling point of water), the saturated vapor pressure of water is approximately equal to what pressure in bar?

Introduction / Context:The normal boiling point of a liquid is defined as the temperature at which its vapor pressure equals the standard atmospheric pressure. For water, this reference point is foundational in steam tables, heat-transfer calculations, and unit operations design. Given Data / Assumptions: Normal boiling point of water is 100°C at 1 atmosphere. Standard atmospheric pressure = 1.013 bar ≈ 101.325 kPa. Concept / Approach:At the normal boiling point, phase equilibrium requires p_sat,water = p_atm. Thus the correct numerical value for saturated vapor pressure at 100°C is the atmospheric pressure in consistent units. Step-by-Step Solution: Recall definition: normal boiling point → p_sat = 1 atm.Convert 1 atm to bar: 1 atm ≈ 1.013 bar.Select 1.013 bar. Verification / Alternative check:Steam tables list p_sat at 100°C as 101.325 kPa = 1.01325 bar, confirming the choice. Why Other Options Are Wrong: 0.1013 bar: That is about 0.1 atm (10.13 kPa), much too low. 10.13 or 101.3 bar: Exceedingly high; 101.3 bar corresponds to about 100 atm. Common Pitfalls:Mixing up kPa and bar; note 1 bar = 100 kPa, not 10 kPa. Final Answer:1.013

Q: Distillation behavior – solutions that boil without composition change In vapor–liquid equilibrium, what are solutions called that distill (boil and condense) without any change in composition of the vapor and liquid phases?

Introduction / Context:An azeotrope is a constant-boiling mixture whose vapor phase has the same composition as the liquid at a given pressure. Azeotropes are central in separation science because they impose limits on simple distillation and often necessitate entrainers or pressure-swing techniques. Given Data / Assumptions: Binary or multicomponent liquid mixtures under a fixed pressure. Well-defined azeotropic composition at the specified pressure. Concept / Approach:At the azeotropic point, y_i = x_i for all key components, so the mixture boils at a constant temperature and composition. Simple batch or differential distillation cannot change the composition past this point, which defines a fundamental separation limit. Ideal, saturated, or supersaturated labels do not capture this constant-composition boiling behavior. Step-by-Step Solution: Define criterion: vapor composition equals liquid composition at boiling.Identify the term that matches this: azeotrope.Select “azeotropic”. Verification / Alternative check:Phase diagrams for ethanol–water show a minimum-boiling azeotrope near 95.6 wt% ethanol at 1 atm; the mixture distills without composition change at the azeotropic point. Why Other Options Are Wrong: Ideal: Refers to Raoult's-law behavior, not constant-composition boiling. Saturated/supersaturated: Concern the amount dissolved, not VLE composition equality. Common Pitfalls:Assuming ideality implies no azeotrope; some near-ideal systems still exhibit weak azeotropy depending on non-ideality and pressure. Final Answer:azeotropic

Q: Quality control of fertilizers – determining urea content from nitrogen assay A urea sample shows an N (nitrogen) content of 42% by weight. Given molecular weight of urea = 60, what is the actual urea content of the sample (mass %)?

Introduction / Context:Fertilizer grade verification often uses nitrogen analysis. For urea, knowing the theoretical nitrogen fraction allows back-calculating the mass fraction of urea in a mixture or degraded sample. This is a routine quality-control calculation in fertilizer plants and agro-retail labs. Given Data / Assumptions: Molecular weight of urea (NH2CONH2) = 60 g/mol. Nitrogen mass per mole of urea = 2 * 14 = 28 g. Measured nitrogen percentage in the sample = 42% by mass. Concept / Approach:Theoretical nitrogen in pure urea is 28/60 by mass = 0.4667 → 46.67%. If a sample analyzes at 42% N, the mass fraction of urea is the ratio of measured N% to theoretical N%, assuming all nitrogen arises from urea. Step-by-Step Solution: Compute theoretical N% in pure urea: (28/60) * 100 = 46.67%.Given sample N% = 42.00%.Mass fraction of urea = 42.00 / 46.67 = 0.900 (approx.).Express as percent: 90% urea by mass. Verification / Alternative check:Take 100 g sample → contains 42 g N → equivalent urea mass = 42 / 0.4667 ≈ 90.0 g, confirming 90%. Why Other Options Are Wrong: 80%, 95%, 98%: Do not match the stoichiometric ratio when using the correct theoretical nitrogen content of urea. Common Pitfalls:Using 50% as rough N content for urea (incorrect); ignoring that all nitrogen detected may not come from urea in adulterated samples (an assumption stated here). Final Answer:90%

Question 1

Dilute solutions — A very dilute solution is prepared by dissolving x1 moles of solute in x2 moles of solvent. The mole fraction of solute is approximately equal to:

Accepted Answer

Introduction / Context:Mole fraction is a fundamental composition variable in thermodynamics and phase equilibria. For very dilute solutions, convenient approximations simplify calculations while maintaining acceptable accuracy. This question tests your ability to derive a first-order approximation for the mole fraction of solute when the solute amount is much smaller than the solvent amount. Given Data / Assumptions: Moles of solute = x1, moles of solvent = x2. Very dilute condition: x1 ≪ x2. Ideal mixing and no volume change on mixing (approximation irrelevant to mole fraction but common context). Concept / Approach: By definition, mole fraction of solute X_solute = x1 / (x1 + x2). Under the dilute condition x1 ≪ x2, the denominator x1 + x2 ≈ x2. Therefore X_solute ≈ x1/x2. This approximation is widely used to linearize expressions in mass transfer and colligative property estimations for trace solutes. Step-by-Step Solution: Start from exact definition: X_solute = x1 / (x1 + x2).Use dilute assumption: x1 + x2 ≈ x2 (since x1 ≪ x2).Therefore: X_solute ≈ x1 / x2.Error in approximation is of order x1/x2 relative to 1, negligible for very small ratios. Verification / Alternative check: Numerical example: x1 = 0.01, x2 = 100 → exact X = 0.01 / 100.01 ≈ 9.999e-5; approximate X ≈ 0.01 / 100 = 1.0e-4. Relative error is <0.001%. Why Other Options Are Wrong: B, D invert or distort the relationship. C approaches 1 and is nonsensical for a dilute solute. E is exact but the question asks for an approximate expression capturing the dilute simplification. Common Pitfalls: Applying the approximation outside the very dilute regime; forgetting that exact mole fraction is bounded between 0 and 1. Final Answer: x1/x2

Question 2

Corresponding states — Gases at the same reduced temperature (Tr) and reduced pressure (Pr) will exhibit which common behavior?

Accepted Answer

Introduction / Context:The principle of corresponding states underpins generalized compressibility charts and equations of state. By normalizing temperature and pressure with respect to critical properties, many gases exhibit similar non-ideal behavior at the same reduced conditions. This allows engineers to estimate deviations from ideality without species-specific data for every case. Given Data / Assumptions: Reduced temperature Tr = T/Tc and reduced pressure Pr = P/Pc. Gases are not in reactive mixtures; behavior is compared at identical Tr and Pr. Acentric factor effects are modest (first-order corresponding states). Concept / Approach: At the same Tr and Pr, many gases show nearly the same compressibility factor Z. Similar Z implies a similar degree of deviation from ideal-gas behavior because Z quantifies PV/(nR*T). While exact equality is not guaranteed (acentric factor ω refines predictions), the “nearly same” behavior forms the basis of generalized charts and correlations. Step-by-Step Solution: Normalize conditions: compute Tr, Pr for each gas.Use generalized Z charts: at the same Tr, Pr, read Z values which cluster for many gases.Infer deviation from ideal gas: deviation ∝ |Z − 1| is similar among gases at same reduced conditions.Hence statements (a) and (b) are both valid. Verification / Alternative check: Comparisons across nitrogen, methane, and argon at identical Tr and Pr show comparable Z values, validating the principle, with small adjustments captured by ω in the Lee–Kesler method. Why Other Options Are Wrong: D rejects a well-founded principle. E concerns critical constants themselves, not reduced-condition behavior. Common Pitfalls: Using corresponding states near phase boundaries without caution; neglecting acentric factor corrections for polar or associating molecules. Final Answer: both (a) and (b)

Question 3

Non-ideality in liquids — The activity coefficient γ of a component in a liquid solution accounts for departure from ideal solution behavior. Which statement is true?

Accepted Answer

Introduction / Context:Real liquid mixtures deviate from Raoult’s-law ideality due to molecular size, shape, and intermolecular forces. The activity coefficient γ modifies the effective concentration (activity) to capture these deviations. Understanding what γ depends on, and what it does not directly represent, is essential for vapor–liquid equilibrium (VLE) and separation process design. Given Data / Assumptions: Activity a_i = γ_i * x_i for liquid-phase component i (at moderate pressures). γ_i depends on composition and temperature (and sometimes pressure weakly). Colligative properties are separate phenomena emergent from solution behavior, not direct definitions of γ. Concept / Approach: In ideal solutions, γ_i = 1. In non-ideal solutions, γ_i ≠ 1 and varies with the mixture composition and temperature because local molecular environments change with composition and thermal energy. While boiling point elevation or freezing point depression can be predicted using activities, γ itself is not “the measure” of those properties; rather, such properties emerge from the full thermodynamic relations. Step-by-Step Solution: Define activity: a_i = γ_i * x_i.Note dependencies: γ_i = f(T, x⃗ , sometimes P).Relate to VLE: y_i * P = x_i * γ_i * P_i^sat (modified Raoult’s law).Conclude: γ_i fundamentally varies with composition (and temperature), not a fixed constant. Verification / Alternative check: Models such as Wilson, NRTL, UNIQUAC compute γ_i as functions of composition and temperature, fitting binary parameters to data. Why Other Options Are Wrong: A/D: Boiling point elevation and freezing point depression are colligative outcomes, not definitions of γ. B: γ typically changes with T. E: γ = 1 only for ideal behavior; many mixtures are non-ideal. Common Pitfalls: Using constant γ over wide composition ranges; ignoring temperature dependence in design calculations. Final Answer: It is a function of the liquid-phase composition

Question 4

Gas mixtures — Equal masses of methane (CH4) and hydrogen (H2) are mixed in an empty container. What fraction of the total pressure is contributed by hydrogen (its partial pressure/total pressure)?

Accepted Answer

Introduction / Context:Dalton’s law of partial pressures states that, for ideal gases, each species’ partial pressure is proportional to its mole fraction in the mixture at a given temperature and total pressure. When mixing equal masses of gases with different molar masses, the lighter gas will contribute more moles—and hence a larger share of the total pressure. This question quantifies that share for H2 and CH4. Given Data / Assumptions: Equal masses of CH4 and H2 are mixed. Ideal-gas behavior; same temperature and total pressure. Molar masses: M_CH4 = 16 g/mol; M_H2 = 2 g/mol. Concept / Approach: Partial pressure fraction = mole fraction for ideal gases. With equal masses, moles are inversely proportional to molar mass. Because hydrogen is eight times lighter than methane, equal masses contain eight times more moles of H2 than CH4, so H2 dominates the mole fraction and therefore the partial pressure. Step-by-Step Solution: Let each gas mass be m grams.Moles of CH4 = m/16; moles of H2 = m/2 = 8*(m/16).Total moles = (m/16) + (8m/16) = 9m/16.Mole fraction of H2 = (8m/16) / (9m/16) = 8/9.Therefore partial pressure of H2 = (8/9)*P_total. Verification / Alternative check: Dalton’s law: p_i = y_i * P_total. The computed mole fraction 8/9 directly sets the partial pressure fraction. Why Other Options Are Wrong: 1/2 suggests equal moles (not the case). 1/9 would correspond to methane’s share. 5/9 or 7/9 are arbitrary and not supported by stoichiometry. Common Pitfalls: Confusing equal masses with equal moles; ignoring the large molar-mass difference which skews mole fractions heavily toward the lighter gas. Final Answer: 8/9

Question 5

Heat capacity integration — Air has Cp ≈ 26.693 + 7.365×10^-3·T (J/mol·K), with T in K. How much heat (kJ) does 1 mol of air give off when cooled at constant pressure from 500°C to −100°C?

Accepted Answer

Introduction / Context:When specific heat depends on temperature, accurate enthalpy changes require integrating Cp over the temperature range. This is common in combustion products, gas cooling, and cryogenic operations. Here, you must perform a straightforward definite integral for Cp(T) over the cooling path to find the heat released by 1 mol of air at constant pressure. Given Data / Assumptions: Cp(T) = 26.693 + 7.36510^-3 * T (J/mol·K). Initial temperature T1 = 500°C = 773.15 K. Final temperature T2 = −100°C = 173.15 K. Constant pressure process; ideal-gas approximation for air. Concept / Approach: For a constant-pressure process, the molar enthalpy change is ΔH = ∫(from T2 to T1) Cp(T) dT if you want the magnitude of heat released (positive). The sign convention for cooling is that the system releases heat: q_p = −∫(from T2 to T1) Cp dT if proceeding from hot to cold. We will compute the magnitude of heat given off. Step-by-Step Solution: Compute ΔT = T1 − T2 = 773.15 − 173.15 = 600.00 K.Integrate Cp(T): ∫Cp dT = ∫[26.693 + 7.36510^-3T] dT = 26.693(ΔT) + (7.36510^-3/2)(T1^2 − T2^2).Calculate the two terms: 26.693600 = 16015.8 J/mol; (7.36510^-3/2)*(773.15^2 − 173.15^2) ≈ 2090.85 J/mol.Sum: 16015.8 + 2090.85 ≈ 18106.65 J/mol = 18.10665 kJ/mol.Rounded to two decimal places: 18.11 kJ/mol of heat released. Verification / Alternative check: A linear Cp(T) gives a quadratic antiderivative. A quick midpoint estimate using Cp at ~473 K also gives a close value, reinforcing the integral result. Why Other Options Are Wrong: 10.73 and 16.15 kJ underestimate the integrated Cp; 18.33 kJ slightly overestimates; 20.50 kJ is too high for the specified range and Cp. Common Pitfalls: Using constant Cp instead of integrating; forgetting unit conversions from J to kJ; mixing Celsius with Kelvin in Cp expressions. Final Answer: 18.11 kJ

Question 6

Diffusion rates in phases – why do gases diffuse faster than liquids? In physical chemistry and transport phenomena, gases typically diffuse much faster than liquids. Which underlying reason best explains this observation in terms of molecular inter…

Accepted Answer

Introduction / Context:Diffusion is the spontaneous mixing of molecules driven by random thermal motion and concentration gradients. A common comparison in transport phenomena is that diffusion coefficients of gases are orders of magnitude larger than those of liquids. Understanding the reason helps in designing absorbers, extractors, and reactors where mass transfer resistance matters. Given Data / Assumptions: Gases have weak intermolecular attractions and large mean free paths. Liquids have much stronger cohesive forces and far smaller molecular mobility. Temperature is comparable in both cases for the qualitative comparison. Concept / Approach:Fick's law relates flux to the diffusion coefficient D. Empirically, D for gases at ambient conditions is about 10^-5 to 10^-4 m^2/s, while for liquids it is typically 10^-10 to 10^-9 m^2/s. The principal cause is not merely molecular mass but the much stronger intermolecular forces and close packing in liquids, which hinder molecular rearrangements and random walks. In gases, molecules are widely spaced; collisions randomize direction yet allow longer displacements between interactions, enabling faster diffusion. Step-by-Step Solution: Identify controlling factor for diffusion: ease of molecular displacement.Recognize liquids exhibit strong cohesive forces (hydrogen bonding, van der Waals, ionic interactions), restricting mobility.Note that gases possess weak interactions and large mean free paths, allowing rapid mixing.Conclude the best explanation is the stronger intermolecular forces in liquids that slow diffusion. Verification / Alternative check:Measured diffusion coefficients support this: D_air for small gases ~ 0.1–1 cm^2/s, whereas D_liquid for solutes in water ~ 10^-5–10^-6 cm^2/s. This five-order magnitude gap cannot be explained by mass differences alone. Why Other Options Are Wrong: Move faster: Liquid molecules generally move slower than gas molecules at the same temperature due to constraints. No definite shape: A macroscopic property unrelated to molecular-level diffusion rate. Are heavier: Molecular mass has some effect, but it is not the dominant reason across phases. Common Pitfalls:Attributing diffusion solely to temperature or mass while ignoring phase structure and cohesion; overlooking that viscosity and packing dominate in liquids. Final Answer:are held together by stronger inter-molecular forces.

Question 7

Acid–base calculations – relation between pH and pOH For an aqueous solution at 25°C with pH = 8, determine the corresponding pOH value using the water ion-product relationship.

Accepted Answer

Introduction / Context:pH and pOH are complementary measures of acidity and basicity in water. At 25°C, their sum is fixed by the ionic product of water. This quick relation is used constantly in environmental engineering, bioprocessing, and analytical chemistry. Given Data / Assumptions: Temperature 25°C (298 K). pH = 8 given. Pure water reference where Kw = 1.0 * 10^-14 at 25°C. Concept / Approach:By definition, pH = -log10[H+] and pOH = -log10[OH-]. The ion product is Kw = [H+]*[OH-] = 1.0 * 10^-14 at 25°C, which leads to pH + pOH = 14. Hence knowing one immediately gives the other by subtraction. Step-by-Step Solution: Use identity: pH + pOH = 14 at 25°C.Substitute pH = 8 → pOH = 14 - 8.Compute pOH = 6. Verification / Alternative check:[H+] = 10^-8 M → [OH-] = Kw / [H+] = 10^-14 / 10^-8 = 10^-6 M → pOH = 6. Consistent. Why Other Options Are Wrong: 1, 7, 10: Do not satisfy pH + pOH = 14 for pH = 8. Common Pitfalls:Using pH + pOH = 14 at temperatures far from 25°C without adjusting Kw; being careless with base-10 logarithms. Final Answer:6

Question 8

Osmotic pressure of dilute solutions – proportionality for a Raoult's-law system For a dilute solution of a non-volatile solute obeying Raoult's law, the osmotic pressure is directly proportional to which single parameter (holding the others constan…

Accepted Answer

Introduction / Context:Osmotic pressure is a colligative property used to estimate molecular weights and design membrane separations. For dilute, ideal solutions, a simple gas-law-like expression connects osmotic pressure to solute amount, solution volume, and absolute temperature. Given Data / Assumptions: Dilute solution of non-volatile solute. Raoult's-law (ideal) behavior is valid. Van't Hoff equation applies. Concept / Approach:The van't Hoff relation is πV = nRT, or π = (n/V) * R * T. Thus π is directly proportional to temperature T at fixed solute amount per unit volume, directly proportional to solute moles n at fixed V and T, and inversely proportional to volume V at fixed n and T. Because the question requests a single proportional factor, the most universal statement among the options is direct proportionality to absolute temperature T. Step-by-Step Solution: Write the relation: π = (n/V) * R * T.Hold n/V constant to isolate the proportional dependence on T.Conclude π ∝ T under the stated ideal, dilute conditions. Verification / Alternative check:Experimentally, increasing temperature at constant concentration raises measured osmotic pressure in line with the gas constant scaling. Why Other Options Are Wrong: Volume of solution: π is inversely proportional to V at fixed n and T, not directly proportional. Moles of solute: True only if V and T are held constant; the prompt asks for a single proportionality, and temperature is the standard statement. None of these: Incorrect because π does scale directly with T for ideal solutions. Common Pitfalls:Failing to state the ceteris paribus condition; forgetting that osmotic pressure depends on concentration and temperature simultaneously. Final Answer:temperature.

Question 9

Water properties – saturated vapor pressure at the normal boiling point At 100°C (the normal boiling point of water), the saturated vapor pressure of water is approximately equal to what pressure in bar?

Accepted Answer

Introduction / Context:The normal boiling point of a liquid is defined as the temperature at which its vapor pressure equals the standard atmospheric pressure. For water, this reference point is foundational in steam tables, heat-transfer calculations, and unit operations design. Given Data / Assumptions: Normal boiling point of water is 100°C at 1 atmosphere. Standard atmospheric pressure = 1.013 bar ≈ 101.325 kPa. Concept / Approach:At the normal boiling point, phase equilibrium requires p_sat,water = p_atm. Thus the correct numerical value for saturated vapor pressure at 100°C is the atmospheric pressure in consistent units. Step-by-Step Solution: Recall definition: normal boiling point → p_sat = 1 atm.Convert 1 atm to bar: 1 atm ≈ 1.013 bar.Select 1.013 bar. Verification / Alternative check:Steam tables list p_sat at 100°C as 101.325 kPa = 1.01325 bar, confirming the choice. Why Other Options Are Wrong: 0.1013 bar: That is about 0.1 atm (10.13 kPa), much too low. 10.13 or 101.3 bar: Exceedingly high; 101.3 bar corresponds to about 100 atm. Common Pitfalls:Mixing up kPa and bar; note 1 bar = 100 kPa, not 10 kPa. Final Answer:1.013

Question 10

Distillation behavior – solutions that boil without composition change In vapor–liquid equilibrium, what are solutions called that distill (boil and condense) without any change in composition of the vapor and liquid phases?

Accepted Answer

Introduction / Context:An azeotrope is a constant-boiling mixture whose vapor phase has the same composition as the liquid at a given pressure. Azeotropes are central in separation science because they impose limits on simple distillation and often necessitate entrainers or pressure-swing techniques. Given Data / Assumptions: Binary or multicomponent liquid mixtures under a fixed pressure. Well-defined azeotropic composition at the specified pressure. Concept / Approach:At the azeotropic point, y_i = x_i for all key components, so the mixture boils at a constant temperature and composition. Simple batch or differential distillation cannot change the composition past this point, which defines a fundamental separation limit. Ideal, saturated, or supersaturated labels do not capture this constant-composition boiling behavior. Step-by-Step Solution: Define criterion: vapor composition equals liquid composition at boiling.Identify the term that matches this: azeotrope.Select “azeotropic”. Verification / Alternative check:Phase diagrams for ethanol–water show a minimum-boiling azeotrope near 95.6 wt% ethanol at 1 atm; the mixture distills without composition change at the azeotropic point. Why Other Options Are Wrong: Ideal: Refers to Raoult's-law behavior, not constant-composition boiling. Saturated/supersaturated: Concern the amount dissolved, not VLE composition equality. Common Pitfalls:Assuming ideality implies no azeotrope; some near-ideal systems still exhibit weak azeotropy depending on non-ideality and pressure. Final Answer:azeotropic

Question 11

Quality control of fertilizers – determining urea content from nitrogen assay A urea sample shows an N (nitrogen) content of 42% by weight. Given molecular weight of urea = 60, what is the actual urea content of the sample (mass %)?

Accepted Answer

Introduction / Context:Fertilizer grade verification often uses nitrogen analysis. For urea, knowing the theoretical nitrogen fraction allows back-calculating the mass fraction of urea in a mixture or degraded sample. This is a routine quality-control calculation in fertilizer plants and agro-retail labs. Given Data / Assumptions: Molecular weight of urea (NH2CONH2) = 60 g/mol. Nitrogen mass per mole of urea = 2 * 14 = 28 g. Measured nitrogen percentage in the sample = 42% by mass. Concept / Approach:Theoretical nitrogen in pure urea is 28/60 by mass = 0.4667 → 46.67%. If a sample analyzes at 42% N, the mass fraction of urea is the ratio of measured N% to theoretical N%, assuming all nitrogen arises from urea. Step-by-Step Solution: Compute theoretical N% in pure urea: (28/60) * 100 = 46.67%.Given sample N% = 42.00%.Mass fraction of urea = 42.00 / 46.67 = 0.900 (approx.).Express as percent: 90% urea by mass. Verification / Alternative check:Take 100 g sample → contains 42 g N → equivalent urea mass = 42 / 0.4667 ≈ 90.0 g, confirming 90%. Why Other Options Are Wrong: 80%, 95%, 98%: Do not match the stoichiometric ratio when using the correct theoretical nitrogen content of urea. Common Pitfalls:Using 50% as rough N content for urea (incorrect); ignoring that all nitrogen detected may not come from urea in adulterated samples (an assumption stated here). Final Answer:90%

Question 12

Gas density at standard conditions – carbon dioxide Express the density of CO2 at normal conditions as kg per normal cubic meter (kg/Nm^3). Choose the correct expression.

Accepted Answer

Introduction / Context:Normal cubic meter (Nm^3) refers to gas volume at standard temperature and pressure. Converting molar properties to mass per unit standard volume is a frequent step in gas metering, emissions reporting, and process calculations. Given Data / Assumptions: Molar mass of CO2 = 44 kg/kmol. Molar volume at STP (approximately) = 22.4 m^3/kmol (for ideal gas at 0°C, 1 atm). Seeking density in kg per Nm^3. Concept / Approach:Density at STP is mass per mole divided by molar volume. For any gas, ρ = M / V_m, where M is molar mass and V_m is molar volume at the stated standard conditions. For CO2: ρ ≈ 44 kg/kmol / 22.4 m^3/kmol. Step-by-Step Solution: Write density formula: ρ = M / V_m.Insert CO2 values: M = 44 kg/kmol; V_m = 22.4 m^3/kmol.Compute expression: ρ = 44/22.4 kg/m^3 ≈ 1.964 kg/Nm^3.Select the expression matching this computation. Verification / Alternative check:Comparing to air (≈28.97/22.4 ≈ 1.29 kg/Nm^3), CO2's higher density is consistent with its higher molar mass. Why Other Options Are Wrong: 44/22400: Uses liters in the denominator but keeps kg units, giving kg per liter-scale denominator numerically wrong for kg/Nm^3. 22.4/44: Inverts the correct ratio, yielding m^3/kg rather than kg/m^3. None of these: Incorrect because 44/22.4 is correct. Common Pitfalls:Mixing L/mol with m^3/mol without consistent unit conversions; forgetting that 1 Nm^3 corresponds to 1 m^3 at standard conditions. Final Answer:44/22.4

Question 13

Estimation rules – what does Kopp's rule calculate? Kopp's rule is a simple additive method used for estimating which thermophysical property of substances from atomic or group contributions?

Accepted Answer

Introduction / Context:When experimental data are unavailable, engineers rely on estimation techniques. Kopp's rule is one of the classic group-additivity methods that provides first-cut estimates useful in material and energy balances and preliminary design. Given Data / Assumptions: Applies primarily to molar heat capacities at constant pressure for solids (and sometimes liquids) at moderate temperatures. Property contributions are additive over atoms or structural groups. Concept / Approach:Kopp's rule states that the molar heat capacity of a compound is approximately the sum of the atomic heat capacities of its constituent elements (or group contributions). While rough, it often yields acceptable estimates near room temperature. It is not designed for conductivity, viscosity, or surface tension, which require different correlations and are more sensitive to structure and state. Step-by-Step Solution: Identify property: molar heat capacity Cp,m.Apply additivity: Cp,m(compound) ≈ Σ Cp,m(atomic contributions).Use as screening estimate when data are scarce. Verification / Alternative check:Comparison with tabulated Cp values shows Kopp's rule typically within reasonable error for many inorganic solids at ambient conditions. Why Other Options Are Wrong: Thermal conductivity, viscosity, surface tension: Not additive in the same simple way; these depend strongly on microstructure and temperature with no universally valid Kopp-style rule. Common Pitfalls:Applying Kopp's rule at high temperatures or to substances with strong anharmonic effects; neglecting phase changes where Cp varies markedly. Final Answer:heat capacity.

Question 14

Liquid–liquid extraction – distribution coefficient (partition ratio) In a ternary system of two immiscible liquids and a solute at equilibrium, the ratio of solute concentrations in the two phases is called the distribution coefficient. On what doe…

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Introduction / Context:The Nernst distribution law governs how a solute partitions between two immiscible solvents. This partition ratio, or distribution coefficient, is fundamental for designing extraction stages and predicting required solvent rates. Given Data / Assumptions: Dilute solutions and ideal behavior assumed initially. Two immiscible liquid phases in contact at equilibrium. Temperature can alter solubility and activity coefficients. Concept / Approach:For truly dilute, non-associating systems, the distribution coefficient can be nearly constant, but in real systems it is affected by solute concentration (non-ideal activity effects, dimerization/association) and temperature (changes in solubility parameters and activity coefficients). Therefore, both temperature and concentration influence K_D to varying degrees, and extraction data are typically reported as a function of both variables. Step-by-Step Solution: Define K_D = C_solute,phase1 / C_solute,phase2 at equilibrium.Recognize activity-coefficient dependence: γ(T, x) changes with composition and temperature.Conclude that both concentration and temperature affect K_D.Select the combined option acknowledging both influences. Verification / Alternative check:Tie-line data in ternary diagrams show K_D varying with composition and shifting with temperature, corroborating the dependence. Why Other Options Are Wrong: Only concentration or only temperature: Incomplete; both exert measurable effects. Neither: Contradicts empirical extraction data and thermodynamic models. Common Pitfalls:Assuming K_D constant over wide ranges; this can mispredict stage counts. Always consult phase-equilibrium data at the operating temperature and composition. Final Answer:both(a) & (b)

Question 15

Thermochemistry – classify the heat change for CS2 formation For the reaction C(s) + 2 S(s) → CS2(l) with a heat change of 104.2 kJ, identify the correct thermochemical term represented by this value.

Accepted Answer

Introduction / Context:Thermochemical data classify heats according to the process: formation, combustion, fusion, solution, etc. Correctly naming a heat effect requires checking both the reactants and the product states relative to their standard reference forms. Given Data / Assumptions: Reaction: C(s) + 2 S(s) → CS2(l). Magnitude of heat change: 104.2 kJ (sign convention omitted in the prompt; classification is requested). Elements in their standard states form one mole of a compound. Concept / Approach:The standard heat of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states at a specified temperature and pressure. Here, solid carbon and solid sulfur are elemental standard forms, and the product is liquid carbon disulfide. Hence the given heat corresponds to the heat (enthalpy) of formation of CS2(l). It is not combustion (which would involve oxygen), fusion (solid ↔ liquid of the same substance), or solution (dissolving a solute into a solvent). Step-by-Step Solution: Compare definition: elements in standard states → compound.Match to given reaction: C(s) + 2S(s) → CS2(l).Therefore classify as heat of formation. Verification / Alternative check:Thermochemical tables list ΔH_f for CS2(l) derived from exactly such a reaction basis (with appropriate sign). Why Other Options Are Wrong: Solution: Would need a solvent and solute. Combustion: Would consume oxygen to form fully oxidized products. Fusion: Phase change of the same substance, not chemical reaction. Common Pitfalls:Misreading the absence of oxygen; assuming any exo/endothermic value implies combustion. Always check whether the process forms the compound from its elements. Final Answer:formation

Question 16

Thermodynamics of solutions:\r When both the solute and the solvent are liquids, the heat effect upon forming the solution is specifically termed the __________.

Accepted Answer

Introduction:Engineers often need to estimate the thermal effects when blending two liquids. The enthalpy change observed when a liquid solute and a liquid solvent are combined at the same initial temperature is a classic topic in solution thermodynamics. Knowing the correct term avoids confusion with gas–liquid or solid–liquid cases and helps in energy balance calculations for mixers and dissolvers. Given Data / Assumptions: Both solute and solvent are in the liquid phase before mixing. No chemical reaction occurs; we consider physical mixing only. The process occurs at constant pressure typical of industrial mixing tanks. Concept / Approach:When two liquids are mixed, the enthalpy change is called the heat of mixing. This differs from heat of hydration (water binding to ions/molecules, usually relevant to dissolving solids or gases in water) and from heat of solvation (a broader term often used for dissolving a solute in a solvent, commonly in gas/solid–liquid contexts). The standard integral heat of solution refers to dissolving a specified amount of solute to form a standard concentration, not necessarily liquid–liquid blending. Hence, for liquid–liquid mixing without reaction, “heat of mixing” is the precise term. Step-by-Step Solution:Identify phases: solute (liquid) + solvent (liquid).Match the proper thermodynamic descriptor: heat of mixing.Eliminate terms associated with hydration/solvation specific to other phase combinations. Verification / Alternative check:Process design texts treat mixer heat duties via enthalpy-of-mixing correlations or excess enthalpy models for liquid mixtures, confirming the terminology. Why Other Options Are Wrong: Heat of hydration/solvation: Emphasize solvent–solute interactions in dissolution (often non-liquid solute). Standard integral heat of solution: A standardized dissolution metric, not generic liquid–liquid mixing. Partial molar enthalpy of dilution: A differential property, not the named overall heat effect. Common Pitfalls:Using “heat of solution” generically for all cases; for two liquids, “heat of mixing” is preferred. Final Answer:heat of mixing.

Question 17

Concentration change on evaporation:\r When an aqueous sugar solution is evaporated (water is removed and sugar is nonvolatile), the molarity of the solution will __________.

Accepted Answer

Introduction:Molarity is a common concentration unit used in material balances and reactor calculations. Understanding how molarity responds to evaporation is crucial in evaporation, crystallization, and syrup concentration operations in the food and biotech industries. Given Data / Assumptions: Solute: sugar (nonvolatile under operating conditions). Solvent: water (volatile and removed by evaporation). Sugar does not decompose; no reactions occur. Concept / Approach:Molarity M is defined as moles of solute per liter of solution. During evaporation, the number of moles of sugar remains constant while the volume of solution decreases as water is removed. Therefore, the ratio moles/volume increases. This remains true for any starting concentration as long as only solvent is removed and temperature-induced volume changes are negligible compared with solvent loss. Step-by-Step Solution:Let initial moles of sugar = n_s (constant during evaporation).Initial volume V1 → final volume V2, where V2 < V1 after evaporating water.M1 = n_s / V1; M2 = n_s / V2.Because V2 < V1, M2 > M1 ⇒ molarity increases. Verification / Alternative check:Mass balance on solute confirms constant solute moles. A quick numerical example (e.g., halving the volume doubles molarity) provides an intuitive check. Why Other Options Are Wrong: Decrease/remain unchanged: Contradict the definition, since volume decreases while solute moles stay fixed. “Either way” or “first decrease then increase”: No such behavior occurs when only solvent is removed and solute is nonvolatile. Common Pitfalls:Confusing molarity with molality. Molality increases even more directly since solvent mass decreases; molarity depends on solution volume. Final Answer:increase

Question 18

Metric micro-scale definition:\r One micron (one micrometre) is numerically equal to which of the following?

Accepted Answer

Introduction:Accurate unit conversion at micro-scales is essential in particle technology, filtration, microfluidics, and microscopy. The micrometre (micron) is a standard SI-derived unit for small lengths such as cell sizes and pore diameters. Given Data / Assumptions: 1 micrometre = 1 μm by definition. We relate μm to meters, centimeters, and millimeters. Base conversions: 1 m = 100 cm = 1000 mm. Concept / Approach:By definition, 1 μm = 10^-6 m. Converting to centimeters: since 1 cm = 10^-2 m, then 10^-6 m = 10^-4 cm. Converting to millimeters: 1 mm = 10^-3 m, hence 1 μm = 10^-3 mm (not 10^-4 mm). Therefore, statements “10^-4 cm” and “10^-6 m” are both correct, while “10^-4 mm” is incorrect. Step-by-Step Solution:Start: 1 μm = 10^-6 m.To cm: 10^-6 m ÷ 10^-2 m/cm = 10^-4 cm.To mm: 10^-6 m ÷ 10^-3 m/mm = 10^-3 mm.Therefore, correct combined choice is both (b) & (c). Verification / Alternative check:Cross-check with a practical value: a 1000 μm object equals 1 mm, consistent with 1 μm = 10^-3 mm. Why Other Options Are Wrong: 10^-4 mm: Off by a factor of 10. 10^-3 mm as a standalone option would be true, but it is not among the original correct pair requested here. Common Pitfalls:Mixing up exponents when converting between cm and mm; always convert via meters to avoid errors. Final Answer:both (b) & (c)

Question 19

“Reduced” properties in corresponding states theory:\r Identify the WRONG statement among the following definitions of reduced temperature, reduced pressure, and reduced volume.

Accepted Answer

Introduction:Reduced properties are central to the law of corresponding states and generalized correlations (e.g., compressibility charts). Precise definitions matter because incorrect temperature scales or reference values lead to serious design errors in phase behavior calculations. Given Data / Assumptions: T_c, P_c, and V_c denote critical temperature, pressure, and molar (or molal) volume. Reduced properties are dimensionless ratios. Thermodynamic temperatures must be in absolute units. Concept / Approach:Reduced temperature is defined as Tr = T / Tc with T and Tc in absolute temperature units (Kelvin or Rankine). Using the Celsius scale is incorrect because Celsius is not an absolute scale; zero Celsius does not represent zero thermal energy. Reduced pressure is Pr = P / Pc. Reduced volume is often Vr = V / Vc (for molar volume) or expressed inversely with compressibility charts. Statements (b) and (c) are correct in spirit; (a) is wrong due to the temperature scale error. Step-by-Step Solution:Check (a): requires absolute temperature → Celsius is wrong.Check (b): ratio of P to Pc is correct.Check (c): ratio of molar (molal) volume to its critical counterpart is correct in definition.Hence, the wrong statement is (a). Verification / Alternative check:Generalized compressibility charts and equations of state explicitly define Tr with absolute scales. Why Other Options Are Wrong: (d) “None” is invalid since (a) is indeed wrong. (e) True: corresponding states correlations employ reduced properties. Common Pitfalls:Plugging Celsius into reduced-temperature formulas causes large numerical errors; always convert to Kelvin. Final Answer:'Reduced temperature' of a substance is the ratio of its existing temperature to its critical temperature, both expressed on celsius scale.

Question 20

Stoichiometry of minerals:\r Calcium carbonate (CaCO3) contains what percentage by mass of calcium (Ca)?

Accepted Answer

Introduction:Weight-percent composition is frequently used in materials balances, fertilizer analysis (e.g., Ca as CaCO3), and hardness calculations. Calcium carbonate is a common reference compound in water treatment and geology. Given Data / Assumptions: Molar masses (approximate): Ca = 40 g/mol, C = 12 g/mol, O = 16 g/mol. Formula: CaCO3. Percent by mass is computed as element mass divided by formula mass times 100. Concept / Approach:The formula mass of CaCO3 is 40 (Ca) + 12 (C) + 3*16 (O) = 100 g/mol (to a good approximation). The calcium mass fraction is then 40/100, or 40%. Using more precise atomic weights (Ca 40.08, C 12.01, O 16.00) gives 40.08/100.09 ≈ 40.0%, confirming the rounded value. Step-by-Step Solution:Compute formula mass: 40 + 12 + 48 = 100 g/mol.Mass of Ca per mole: 40 g.Percent Ca = (40 / 100) * 100% = 40%. Verification / Alternative check:Exact atomic weights change the result only in the second decimal place, so 40% is a standard taught value. Why Other Options Are Wrong: 48%: This is close to the oxygen fraction per atom but not Ca. 96% or 12%: Physically impossible based on the stoichiometry. 56%: No element in CaCO3 has this fraction. Common Pitfalls:Forgetting there are three oxygens or mis-adding the atomic masses. Final Answer:40

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