Q: A sample of gas occupies 2.50 L at 25°C and 1.00 atm. If the temperature is increased to 75°C at constant pressure, what will be the new volume of the gas?

$$V_2$$ = $$V_1$$ \times \frac{$$T_2$$}{$$T_1$$} = 2.50\ \mathrm{L} \times \frac{348.15\ \mathrm{K}}{298.15\ \mathrm{K}} = 2.92\ \mathrm{L}$$

Q: A 0.300 g sample of a liquid is vaporized at 150.0°C. The vapor occupies a volume of 180.0 mL at 0.998 atm. What is the molar mass of the liquid? (R = 0.08206 L·atm·mol⁻¹·K⁻¹)

$$M = \frac{mRT}{PV} = \frac{0.300\ \mathrm{g} \times 0.08206\ \mathrm{L\cdot atm\cdot $$mol^{-1}$$\cdot K$$^{-1}$$} \times 423.15\ \mathrm{K}}{0.998\ \mathrm{atm} \times 0.1800\ \mathrm{L}} = 62.0\ \mathrm{g\cdot $$mol^{-1}$$}$$

Q: A mixture of gases contains 0.75 mol N₂, 0.22 mol O₂, and 0.50 mol CO₂. If the total pressure is 1.86 atm, what is the partial pressure of O₂?

$$P_{O_2$} = X_{O_2$} \times P_{\text{total}} = \frac{0.22}{0.75+0.22+0.50} \times 1.86\ \mathrm{atm} = 0.33\ \mathrm{atm}$$

Q: A gas has a density of 0.572 g/L at 90.0°C and 380.0 mm Hg pressure. What is the molar mass of the gas? (R = 0.08206 L·atm·mol⁻¹·K⁻¹)

$$M = \frac{dRT}{P} = \frac{0.572\ \mathrm{g/L} \times 0.08206\ \mathrm{L\cdot atm\cdot $$mol^{-1}$$\cdot K$$^{-1}$$} \times 363.15\ \mathrm{K}}{0.500\ \mathrm{atm}} = 34.1\ \mathrm{g\cdot $$mol^{-1}$$}$$

Q: A sample of gas at 338 torr and 45°C occupies 206 mL. What is the molar mass if the mass of the gas is 0.206 g? (R = 0.08206 L·atm·mol⁻¹·K⁻¹)

$$M = \frac{mRT}{PV} = \frac{0.206\ \mathrm{g} \times 0.08206\ \mathrm{L\cdot atm\cdot $$mol^{-1}$$\cdot K$$^{-1}$$} \times 318.15\ \mathrm{K}}{0.445\ \mathrm{atm} \times 0.206\ \mathrm{L}} = 60.0\ \mathrm{g\cdot $$mol^{-1}$$}$$

Question 1

A sample of gas occupies 2.50 L at 25°C and 1.00 atm. If the temperature is increased to 75°C at constant pressure, what will be the new volume of the gas?

Accepted Answer

$$V_2$$ = $$V_1$$ 	imes \frac{$$T_2$$}{$$T_1$$} = 2.50\ \mathrm{L} 	imes \frac{348.15\ \mathrm{K}}{298.15\ \mathrm{K}} = 2.92\ \mathrm{L}$$

Question 2

Which of the following statements best describes Dalton's Law of Partial Pressures?

Accepted Answer

The total pressure of a mixture of gases is equal to the sum of the partial pressures of each individual gas.

Question 3

A 0.300 g sample of a liquid is vaporized at 150.0°C. The vapor occupies a volume of 180.0 mL at 0.998 atm. What is the molar mass of the liquid? (R = 0.08206 L·atm·mol⁻¹·K⁻¹)

Accepted Answer

$$M = \frac{mRT}{PV} = \frac{0.300\ \mathrm{g} 	imes 0.08206\ \mathrm{L\cdot atm\cdot $$mol^{-1}$$\cdot K$$^{-1}$$} 	imes 423.15\ \mathrm{K}}{0.998\ \mathrm{atm} 	imes 0.1800\ \mathrm{L}} = 62.0\ \mathrm{g\cdot $$mol^{-1}$$}$$

Question 4

Which of the following best explains why lighter gases diffuse faster than heavier gases?

Accepted Answer

According to Graham's law, the rate of diffusion is inversely proportional to the square root of the molar mass: $$	ext{rate} \propto \frac{1}{\sqrt{M}}$$

Question 5

A mixture of gases contains 0.75 mol N₂, 0.22 mol O₂, and 0.50 mol CO₂. If the total pressure is 1.86 atm, what is the partial pressure of O₂?

Accepted Answer

$$P_{O_2$} = X_{O_2$} 	imes P_{	ext{total}} = \frac{0.22}{0.75+0.22+0.50} 	imes 1.86\ \mathrm{atm} = 0.33\ \mathrm{atm}$$

Question 6

Which of the following is the correct value for the molar volume of an ideal gas at STP (0°C, 1 atm)?

Accepted Answer

$$22.4\ \mathrm{L}$$

Question 7

A 1.00 L sample of gas is collected at 25°C and 1.25 atm. What is the pressure of the gas (in mm Hg) at 200.0°C if the volume is 4.00 L?

Accepted Answer

$$P_2$$ = \frac{$$P_1$$ $$V_1$$ $$T_2$$}{$$T_1$$ $$V_2$$} = \frac{1.25\ \mathrm{atm} 	imes 1.00\ \mathrm{L} 	imes 473.15\ \mathrm{K}}{298.15\ \mathrm{K} 	imes 4.00\ \mathrm{L}} = 0.50\ \mathrm{atm} = 380\ \mathrm{mm\ Hg}$$

Question 8

Which of the following best describes the function of a barometer?

Accepted Answer

A barometer measures atmospheric pressure by balancing the weight of a column of mercury against the atmospheric pressure.

Question 9

A gas has a density of 0.572 g/L at 90.0°C and 380.0 mm Hg pressure. What is the molar mass of the gas? (R = 0.08206 L·atm·mol⁻¹·K⁻¹)

Accepted Answer

$$M = \frac{dRT}{P} = \frac{0.572\ \mathrm{g/L} 	imes 0.08206\ \mathrm{L\cdot atm\cdot $$mol^{-1}$$\cdot K$$^{-1}$$} 	imes 363.15\ \mathrm{K}}{0.500\ \mathrm{atm}} = 34.1\ \mathrm{g\cdot $$mol^{-1}$$}$$

Question 10

Which of the following is NOT an assumption of the ideal gas law?

Accepted Answer

Gas particles have significant volume compared to the container.

Question 11

A sample of gas at 338 torr and 45°C occupies 206 mL. What is the molar mass if the mass of the gas is 0.206 g? (R = 0.08206 L·atm·mol⁻¹·K⁻¹)

Accepted Answer

$$M = \frac{mRT}{PV} = \frac{0.206\ \mathrm{g} 	imes 0.08206\ \mathrm{L\cdot atm\cdot $$mol^{-1}$$\cdot K$$^{-1}$$} 	imes 318.15\ \mathrm{K}}{0.445\ \mathrm{atm} 	imes 0.206\ \mathrm{L}} = 60.0\ \mathrm{g\cdot $$mol^{-1}$$}$$

Question 12

Which of the following best describes the relationship between pressure and volume for a fixed amount of gas at constant temperature (Boyle's Law)?

Accepted Answer

$$P_1$$ $$V_1$$ = $$P_2$$ $$V_2$$; pressure and volume are inversely proportional.

Question 13

A mixture of CO and Kr weighs 3.60 g and exerts a pressure of 0.70 atm in its container. Since Kr is expensive, you wish to recover it from the mixture. After the CO is completely removed, the pressure drops to 0.20 atm. How many grams of Kr can you recover? (Molar mass of Kr = 83.8 g/mol)

Accepted Answer

$$P_{	ext{Kr}} = 0.20\ \mathrm{atm}$$; 
$$\frac{P_{	ext{Kr}}}{P_{	ext{total}}} = \frac{n_{	ext{Kr}}}{n_{	ext{total}}}$$; 
$$n_{	ext{Kr}} = \frac{P_{	ext{Kr}}}{P_{	ext{total}}} 	imes n_{	ext{total}}$$; 
$$	ext{Let } n_{	ext{total}} = \frac{3.60\ \mathrm{g}}{M_{	ext{CO}} + M_{	ext{Kr}}}$$; 
	ext{Solve for } n_{	ext{Kr}}$$; 
$$	ext{Final answer: } 0.72\ \mathrm{g}$

Question 14

Which of the following is the correct expression for the mole fraction ($$X_A) of $$component A in a mixture?

Accepted Answer

$$X_A = \frac{n_A}{n_{	ext{total}}}$$

Question 15

A sample of gas at 1.00 atm and 0°C has a volume of 22.4 L. If the pressure is doubled and the temperature is increased to 273°C, what is the new volume?

Accepted Answer

$$V_2$$ = $$V_1$$ 	imes \frac{$$P_1$$}{$$P_2$$} 	imes \frac{$$T_2$$}{$$T_1$$} = 22.4\ \mathrm{L} 	imes \frac{1.00}{2.00} 	imes \frac{546.15\ \mathrm{K}}{273.15\ \mathrm{K}} = 22.4\ \mathrm{L}$$

Question 16

Which of the following best describes diffusion?

Accepted Answer

Diffusion is the spreading of a gas in an open space where it mixes with another gas.

Question 17

A sample of gas has a mass of 207 mL at 699 mm Hg and 45°C. What is the mass (in grams) of the sample if its molar mass is 98.0 g/mol? (R = 0.08206 L·atm·mol⁻¹·K⁻¹)

Accepted Answer

$$n = \frac{PV}{RT} = \frac{0.920\ \mathrm{atm} 	imes 0.207\ \mathrm{L}}{0.08206\ \mathrm{L\cdot atm\cdot $$mol^{-1}$$\cdot K$$^{-1}$$} 	imes 318.15\ \mathrm{K}} = 0.0074\ \mathrm{mol}$$; 
$$m = n 	imes M = 0.0074\ \mathrm{mol} 	imes 98.0\ \mathrm{g/mol} = 0.73\ \mathrm{g}$$

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