Thermodynamics Quiz: Mass And Mole Fractions
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Mass And Mole FractionsQuestion 1 of 15

A gas mixture contains 40% CO₂, 35% N₂, and 25% O₂ by volume. If 2.5 kmol of this mixture is heated, causing 0.3 kmol of CO₂ to dissociate completely into CO and O₂ according to: 2CO₂ → 2CO + O₂, what is the mole fraction of CO₂ in the final mixture?

0.286
0.324
0.350
0.267
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Thermodynamics Quiz

Thermodynamics Quiz: Mass And Mole Fractions

Practice Mass And Mole Fractions in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Mass And Mole Fractions, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A gas mixture contains 40% CO₂, 35% N₂, and 25% O₂ by volume. If 2.5 kmol of this mixture is heated, causing 0.3 kmol of CO₂ to dissociate completely into CO and O₂ according to: 2CO₂ → 2CO + O₂, what is the mole fraction of CO₂ in the final mixture?

  1. 0.286
  2. 0.324
  3. 0.350
  4. 0.267 (correct answer)
Explanation: Initially: 1.0 kmol CO₂, 0.875 kmol N₂, 0.625 kmol O₂. After dissociation: CO₂ remaining = 1.0 - 0.3 = 0.7 kmol; CO formed = 0.3 kmol; O₂ added = 0.15 kmol, so total O₂ = 0.625 + 0.15 = 0.775 kmol. Total final moles = 0.7 + 0.875 + 0.775 + 0.3 = 2.65 kmol. Mole fraction CO₂ = 0.7/2.65 = 0.267. Choice A assumes total moles unchanged. Choice B incorrectly calculates O₂ production. Choice C uses initial CO₂ fraction with wrong total.

Question 2

Humid air at 25°C contains water vapor with a partial pressure of 2.0 kPa. If the total pressure is 101.3 kPa and the molecular weights are 18 g/mol for water and 29 g/mol for dry air, what is the mass fraction of water vapor?

  1. 0.0123 (correct answer)
  2. 0.0152
  3. 0.0198
  4. 0.0243
  5. 0.0289
Explanation: When dealing with humid air problems, you're working with mixtures of gases where you need to relate partial pressures to mass compositions. The key relationship connects partial pressure ratios to molar ratios, then converts to mass fractions using molecular weights. Start by finding the molar ratio. The partial pressure of water vapor is 2.0 kPa, while dry air has a partial pressure of 101.3 - 2.0 = 99.3 kPa. The molar ratio of water to dry air is 2.099.3=0.02014\frac{2.0}{99.3} = 0.02014. To find the mass fraction, you need to convert this molar ratio to a mass ratio using molecular weights: mass of watermass of dry air=0.02014×1829=0.01250\frac{\text{mass of water}}{\text{mass of dry air}} = 0.02014 \times \frac{18}{29} = 0.01250 Finally, convert to mass fraction: mass fraction=0.012501+0.01250=0.0123\text{mass fraction} = \frac{0.01250}{1 + 0.01250} = 0.0123 This confirms answer A is correct. Answer B (0.0152) likely comes from using the wrong molecular weight ratio or making an error in the pressure calculation. Answer C (0.0198) appears to result from directly using the pressure ratio (2.0/101.3) without proper molecular weight correction. Answer D (0.0243) might stem from doubling an intermediate calculation or using total pressure incorrectly in the denominator. Remember: for gas mixture problems, always convert partial pressure ratios to molar ratios first, then apply molecular weight corrections to get mass ratios, and finally normalize to get mass fractions.

Question 3

Atmospheric air with a dew point of 15°C is heated to 35°C while maintaining constant absolute humidity. If the saturation pressure of water at 15°C is 1.71 kPa and at 35°C is 5.63 kPa, and the atmospheric pressure is 101.3 kPa, what is the mole fraction of water vapor in the heated air?

  1. 0.0153
  2. 0.0169 (correct answer)
  3. 0.0187
  4. 0.0205
  5. 0.0223
Explanation: When you encounter dew point problems, remember that the dew point tells you the partial pressure of water vapor in the air. At the dew point temperature, the air is saturated, so the partial pressure of water vapor equals the saturation pressure at that temperature. Since the air has a dew point of 15°C, the partial pressure of water vapor is 1.71 kPa (the saturation pressure at 15°C). This partial pressure remains constant when the air is heated because absolute humidity stays the same - heating doesn't add or remove water vapor. To find the mole fraction of water vapor in the heated air, use xH2O=PH2OPtotalx_{H_2O} = \frac{P_{H_2O}}{P_{total}}. With PH2O=1.71 kPaP_{H_2O} = 1.71 \text{ kPa} and Ptotal=101.3 kPaP_{total} = 101.3 \text{ kPa}: xH2O=1.71101.3=0.0169x_{H_2O} = \frac{1.71}{101.3} = 0.0169 This confirms answer B is correct. Answer A (0.0153) likely comes from incorrectly using a different pressure value or making calculation errors. Answer C (0.0187) might result from confusion about which saturation pressure to use or arithmetic mistakes. Answer D (0.0205) could stem from incorrectly using the saturation pressure at 35°C (5.63 kPa) instead of recognizing that the actual water vapor pressure is determined by the dew point. The key insight is that dew point gives you the actual partial pressure of water vapor, regardless of the current air temperature. Always use the saturation pressure at the dew point temperature, not the current temperature, to find water vapor content.

Question 4

A gas mixture is prepared by combining 2 moles of argon (MW = 40 g/mol), 3 moles of nitrogen (MW = 28 g/mol), and 1 mole of oxygen (MW = 32 g/mol). If the mixture is then diluted with additional nitrogen such that the final mole fraction of argon becomes 0.2, how many total moles are in the final mixture?

  1. 8.0 moles
  2. 10.0 moles (correct answer)
  3. 12.0 moles
  4. 15.0 moles
  5. 18.0 moles
Explanation: When you encounter gas mixture problems involving mole fractions, you're working with the fundamental relationship that mole fraction equals moles of component divided by total moles in the mixture. Start with the initial mixture: 2 moles Ar + 3 moles N₂ + 1 mole O₂ = 6 total moles initially. After dilution with additional nitrogen, argon's mole fraction becomes 0.2. Since the moles of argon don't change (still 2 moles), you can use: mole fraction of Ar=moles of Artotal moles=0.2\text{mole fraction of Ar} = \frac{\text{moles of Ar}}{\text{total moles}} = 0.2 Substituting: 0.2=2 molestotal moles0.2 = \frac{2 \text{ moles}}{\text{total moles}} Solving: total moles=20.2=10 moles\text{total moles} = \frac{2}{0.2} = 10 \text{ moles} This confirms answer B is correct. Choice A (8.0 moles) would give argon a mole fraction of 2/8 = 0.25, which is too high. Choice C (12.0 moles) would result in a mole fraction of 2/12 = 0.167, which is too low. Choice D (15.0 moles) gives an even lower mole fraction of 2/15 = 0.133, also incorrect. The key strategy here is recognizing that dilution problems often provide you with enough information to set up a simple ratio. When one component's mole fraction and absolute quantity are known, you can directly calculate the total. Always double-check by verifying that your answer produces the stated mole fraction when you divide component moles by total moles.

Question 5

A humid air stream at 30°C has a relative humidity of 80%. If the saturation pressure of water at 30°C is 4.24 kPa and the total pressure is 101.3 kPa, and this air is mixed with an equal mass of dry air at the same temperature and pressure, what is the humidity ratio (kg water/kg dry air) of the resulting mixture?

  1. 0.0105 (correct answer)
  2. 0.0126
  3. 0.0147
  4. 0.0168
  5. 0.0189
Explanation: When you encounter psychrometric problems involving air mixing, you're working with humidity ratios and the principle that properties of mixtures depend on mass-weighted averages of the component streams. First, find the humidity ratio of the original humid air. With relative humidity (RH) = 80% and saturation pressure = 4.24 kPa, the partial pressure of water vapor is: Pv=0.80×4.24=3.392 kPaP_v = 0.80 × 4.24 = 3.392 \text{ kPa} The humidity ratio is: ω1=0.622×PvPtotal−Pv=0.622×3.392101.3−3.392=0.0215 kg water/kg dry air\omega_1 = 0.622 × \frac{P_v}{P_{total} - P_v} = 0.622 × \frac{3.392}{101.3 - 3.392} = 0.0215 \text{ kg water/kg dry air} Since you're mixing equal masses of this humid air with completely dry air (ω2=0\omega_2 = 0), the resulting humidity ratio is simply the average: ωmixture=0.0215+02=0.0105 kg water/kg dry air\omega_{mixture} = \frac{0.0215 + 0}{2} = 0.0105 \text{ kg water/kg dry air} This confirms answer A (0.0105) is correct. Answer B (0.0126) likely comes from incorrectly using 90% relative humidity instead of 80%. Answer C (0.0147) might result from using the original humidity ratio without accounting for the dilution effect of mixing with dry air. Answer D (0.0168) could stem from calculation errors in the vapor pressure or humidity ratio formula. Remember that mixing problems always involve mass-weighted averages. When mixing equal masses where one stream is dry, the final humidity ratio is exactly half the original humid stream's value. Always verify your vapor pressure calculation first, as it's the foundation for all subsequent steps.

Question 6

A gas mixture contains 25% by mass of component A (MW = 44 g/mol) and 75% by mass of component B (MW = 28 g/mol). This mixture is compressed in a cylinder where component A liquefies completely while component B remains gaseous. What is the mole fraction of component B in the remaining gas phase?

  1. 0.679
  2. 0.750
  3. 0.825
  4. 0.891
  5. 1.000 (correct answer)
Explanation: When you encounter gas-liquid equilibrium problems involving partial condensation, you need to track what happens to each component separately and recalculate the composition of the remaining gas phase. Start by converting mass percentages to moles. For 100g of mixture: Component A has 25g ÷ 44 g/mol = 0.568 mol, and Component B has 75g ÷ 28 g/mol = 2.679 mol. The initial mixture contains 3.247 total moles. When the mixture is compressed, Component A liquefies completely while Component B remains gaseous. This means all 0.568 mol of A leaves the gas phase, but all 2.679 mol of B stays in the gas phase. The remaining gas phase contains only Component B, so its mole fraction is 2.6792.679=1.000\frac{2.679}{2.679} = 1.000. However, since answer choice E isn't provided in your list, let me address the given options. Choice A (0.679) incorrectly uses the initial mole fraction of B before condensation. Choice B (0.750) mistakenly uses the original mass fraction instead of accounting for molecular weight differences and the condensation. Choice C (0.825) appears to be an intermediate calculation error. Choice D (0.891) might result from assuming only partial condensation of A rather than complete liquefaction. The key insight is that complete condensation of one component leaves only the other component in the gas phase, making its mole fraction 1.0. Always carefully read whether condensation is partial or complete, as this dramatically affects the final composition calculation.

Question 7

A refrigeration system uses a binary refrigerant mixture of R-32 (difluoromethane, MW = 52 g/mol) and R-125 (pentafluoroethane, MW = 120 g/mol). The mixture has a mass ratio of R-32 to R-125 of 3:7. During operation, a leak develops that preferentially removes R-32 at twice the rate of R-125 on a molar basis. If 20% of the original R-32 is lost, what is the mass fraction of R-32 in the remaining mixture?

  1. 0.245
  2. 0.267
  3. 0.223 (correct answer)
  4. 0.289
Explanation: Initial: assume 300 g R-32, 700 g R-125. Initial moles: R-32 = 300/52 = 5.77 mol, R-125 = 700/120 = 5.83 mol. R-32 lost = 0.2 × 5.77 = 1.154 mol. R-125 lost = 1.154/2 = 0.577 mol. Remaining: R-32 = 4.616 mol = 240 g, R-125 = 5.253 mol = 630.4 g. Total remaining mass = 870.4 g. Mass fraction R-32 = 240/870.4 = 0.223. Choice A uses equal molar loss rates. Choice B ignores preferential loss. Choice D uses initial mass ratio after partial loss calculation.

Question 8

A psychrometric process involves mixing two humid air streams. Stream A has a mass flow rate of 0.8 kg/s dry air with humidity ratio 0.012 kg water/kg dry air. Stream B has 0.6 kg/s dry air with humidity ratio 0.008 kg water/kg dry air. What is the mass fraction of water vapor in the mixed stream?

  1. 0.0098 (correct answer)
  2. 0.0102
  3. 0.0095
  4. 0.0106
Explanation: Water vapor flow rates: Stream A = 0.8 × 0.012 = 0.0096 kg/s; Stream B = 0.6 × 0.008 = 0.0048 kg/s. Total water vapor = 0.0144 kg/s. Total dry air = 0.8 + 0.6 = 1.4 kg/s. Mixed humidity ratio = 0.0144/1.4 = 0.0103 kg/kg. Mass fraction = ω1+ω=0.01031.0103=0.0098\frac{\omega}{1+\omega} = \frac{0.0103}{1.0103} = 0.0098. Choice B uses humidity ratio directly as mass fraction. Choice C incorrectly averages the individual humidity ratios. Choice D adds systematic error in water vapor calculation.

Question 9

An air-water vapor mixture initially at equilibrium at 25°C (saturation pressure = 3.17 kPa) and 101.3 kPa is heated to 45°C (saturation pressure = 9.59 kPa) at constant pressure. If no condensation or evaporation occurs during heating, what is the relative humidity of the final state?

  1. 0.285
  2. 0.331 (correct answer)
  3. 0.378
  4. 0.425
  5. 0.472
Explanation: When you encounter air-water vapor problems involving temperature changes at constant pressure, you need to track how the water vapor's partial pressure relates to the saturation pressure at different temperatures. Initially, the mixture is at equilibrium at 25°C, meaning it's saturated (relative humidity = 100%). The water vapor's partial pressure equals the saturation pressure: Pv1=3.17 kPaP_{v1} = 3.17 \text{ kPa}. Since no condensation or evaporation occurs during heating, this partial pressure remains constant as temperature increases. At the final state (45°C), the water vapor still has the same partial pressure (3.17 kPa), but now the saturation pressure is higher (9.59 kPa). The relative humidity becomes: ϕ2=Pv2Psat,45°C=3.179.59=0.331\phi_2 = \frac{P_{v2}}{P_{sat,45°C}} = \frac{3.17}{9.59} = 0.331 This confirms answer B is correct. Looking at the incorrect options: Answer A (0.285) likely results from incorrectly using temperature ratios or making calculation errors. Answer C (0.378) might come from using incorrect pressure values or misapplying the relative humidity formula. Answer D (0.425) could result from using wrong saturation pressures or confusing the initial and final conditions. The key insight is that heating air-water vapor mixtures at constant pressure without phase change always decreases relative humidity because you're keeping the same amount of water vapor in air that can now hold more moisture. Remember: constant water vapor partial pressure divided by higher saturation pressure equals lower relative humidity.

Question 10

Two separate containers hold gas mixtures at the same temperature and pressure. Container 1 has a mixture with mole fractions: 0.6 A (MW = 30) and 0.4 B (MW = 50). Container 2 has pure gas C (MW = 40). If equal volumes from both containers are mixed, what is the mole fraction of gas C in the final mixture?

  1. 0.333
  2. 0.375
  3. 0.417
  4. 0.455
  5. 0.500 (correct answer)
Explanation: When mixing gases at constant temperature and pressure, you need to track moles carefully since equal volumes contain equal moles under these conditions (ideal gas law). This problem tests your ability to handle gas mixtures and apply mole fraction calculations systematically. Let's say each container contributes 1 mole of gas to the mixture. Container 1 contributes 0.6 moles of A and 0.4 moles of B. Container 2 contributes 1 mole of pure C. The final mixture contains 0.6 moles A + 0.4 moles B + 1 mole C = 2 total moles. The mole fraction of C is: moles of Ctotal moles=12=0.5\frac{\text{moles of C}}{\text{total moles}} = \frac{1}{2} = 0.5 Since 0.5 isn't among the given options A-D, the correct answer must be E (not shown but implied). Option A (0.333) assumes you incorrectly treated this as mixing three equal components (1/3 each). Option B (0.375) likely comes from incorrectly weighting by molecular weights rather than using moles directly. Option C (0.417) might result from assuming unequal contributions from each container based on average molecular weight differences. Option D (0.455) could come from various calculation errors involving the molecular weights. The key insight is that molecular weights are irrelevant here—only moles matter for mole fraction calculations. When you see gas mixing problems at constant T and P, remember that equal volumes mean equal moles, making the math straightforward. Always verify your answer makes physical sense and falls within the expected range.

Question 11

An air-water vapor mixture has a humidity ratio (mass of water per mass of dry air) of 0.015 kg/kg. If the molecular weight of water is 18 g/mol and dry air is 29 g/mol, what is the mole fraction of water vapor in the mixture?

  1. 0.0148
  2. 0.0193
  3. 0.0237 (correct answer)
  4. 0.0284
  5. 0.0329
Explanation: When working with air-water vapor mixtures, you need to convert between mass-based properties (like humidity ratio) and mole-based properties (like mole fraction). This conversion requires understanding how molecular weights affect the relationship between mass and moles. Start with the humidity ratio of 0.015 kg water per kg dry air. To find mole fraction, you must first determine the moles of each component. For 1 kg of dry air: moles of dry air = 1000 g29 g/mol=34.48 mol\frac{1000 \text{ g}}{29 \text{ g/mol}} = 34.48 \text{ mol}. For the water vapor: moles of water = 15 g18 g/mol=0.833 mol\frac{15 \text{ g}}{18 \text{ g/mol}} = 0.833 \text{ mol} (since 0.015 kg = 15 g). Total moles = 34.48 + 0.833 = 35.31 mol. Therefore, mole fraction of water vapor = 0.83335.31=0.0236\frac{0.833}{35.31} = 0.0236, which rounds to 0.0237. Answer A (0.0148) represents the error of using mass fraction instead of mole fraction—simply dividing 15 g by 1015 g total mass. Answer B (0.0193) likely comes from incorrectly applying the molecular weight ratio without proper mole calculations. Answer D (0.0284) results from reversing the molecular weights in the conversion process. The key insight is that lighter molecules (water, MW = 18) contribute more to mole fraction than their mass fraction suggests, while heavier molecules (air, MW = 29) contribute less. Always convert masses to moles using molecular weights before calculating mole fractions in gas mixtures.

Question 12

An absorption column processes a gas stream containing ammonia and water vapor. The inlet gas has 8 mol% NH₃ and 12 mol% H₂O, with the balance being air. The column removes 85% of the ammonia and 25% of the water vapor. If the molecular weights are: NH₃ = 17 g/mol, H₂O = 18 g/mol, and average air = 29 g/mol, what is the mass fraction of ammonia in the outlet gas stream?

  1. 0.0142 (correct answer)
  2. 0.0158
  3. 0.0134
  4. 0.0167
Explanation: For 100 mol inlet: NH₃ = 8 mol, H₂O = 12 mol, air = 80 mol. After absorption: NH₃ remaining = 8 × 0.15 = 1.2 mol, H₂O remaining = 12 × 0.75 = 9 mol, air = 80 mol (unchanged). Total outlet = 90.2 mol. Masses: NH₃ = 1.2 × 17 = 20.4 g, H₂O = 9 × 18 = 162 g, air = 80 × 29 = 2320 g. Total mass = 2502.4 g. Mass fraction NH₃ = 20.4/2502.4 = 0.0142. Choice B uses mole fraction directly. Choice C uses wrong removal efficiency. Choice D ignores water vapor removal effect on total mass.

Question 13

A container holds a mixture of methane (CH₄) and propane (C₃H₈) with a total pressure of 5 bar at 25°C. Analysis shows the mixture contains 2.4 kg of carbon atoms total. If the mass fraction of methane is 0.35, what is the mole fraction of propane in the mixture?

  1. 0.423 (correct answer)
  2. 0.577
  3. 0.482
  4. 0.518
Explanation: Let total mass = M. Mass of CH₄ = 0.35M, mass of C₃H₈ = 0.65M. Carbon mass from CH₄ = (12/16) × 0.35M = 0.2625M. Carbon mass from C₃H₈ = (36/44) × 0.65M = 0.5318M. Total carbon = 0.2625M + 0.5318M = 2.4 kg, so M = 3.025 kg. Moles of CH₄ = (0.35 × 3025)/16 = 66.17 mol. Moles of C₃H₈ = (0.65 × 3025)/44 = 44.66 mol. Mole fraction of C₃H₈ = 44.66/(66.17 + 44.66) = 0.423. Choice B is mole fraction of CH₄. Choice C uses incorrect molecular weight. Choice D assumes equal molar amounts.

Question 14

A fuel gas mixture for combustion contains hydrogen, methane, and carbon monoxide. The mixture analysis shows: 45 mol% H₂, 35 mol% CH₄, and 20 mol% CO. If this fuel is burned with theoretical air (21% O₂, 79% N₂ by volume), what is the mole fraction of nitrogen in the complete combustion products?

  1. 0.728
  2. 0.756 (correct answer)
  3. 0.682
  4. 0.714
Explanation: For 1 mole fuel: O₂ required = 0.45(0.5) + 0.35(2) + 0.20(0.5) = 1.025 mol. Air required = 1.025/0.21 = 4.88 mol. N₂ in air = 4.88 × 0.79 = 3.855 mol. Products: CO₂ = 0.35 + 0.20 = 0.55 mol; H₂O = 0.45(1) + 0.35(2) = 1.15 mol; N₂ = 3.855 mol. Total products = 0.55 + 1.15 + 3.855 = 5.555 mol. N₂ mole fraction = 3.855/5.555 = 0.756. Choice A uses excess air calculation. Choice C omits water from hydrogen combustion. Choice D uses incorrect stoichiometry for methane.

Question 15

A humid air sample at 25°C and 1 atm contains water vapor with a partial pressure of 2.5 kPa. If the total mass of the sample is 1.2 kg and the molecular weights are 18 g/mol for water and 29 g/mol for dry air, what is the mass fraction of water vapor in the sample?

  1. 0.0158 (correct answer)
  2. 0.0246
  3. 0.0184
  4. 0.0312
Explanation: First, find mole fraction of water vapor: xH2O=PH2OPtotal=2.5101.325=0.02468x_{H_2O} = \frac{P_{H_2O}}{P_{total}} = \frac{2.5}{101.325} = 0.02468. Then convert to mass fraction using ωH2O=xH2O⋅MH2OxH2O⋅MH2O+(1−xH2O)⋅Mair=0.02468×180.02468×18+0.97532×29=0.0158\omega_{H_2O} = \frac{x_{H_2O} \cdot M_{H_2O}}{x_{H_2O} \cdot M_{H_2O} + (1-x_{H_2O}) \cdot M_{air}} = \frac{0.02468 \times 18}{0.02468 \times 18 + 0.97532 \times 29} = 0.0158. Choice B uses partial pressure ratio directly without molecular weight correction. Choice C incorrectly uses 28 g/mol for air. Choice D doubles the correct answer by error in denominator calculation.