1. What is the volume of 0.900 moles of an ideal gas at 25.0° C and a pressure of 950.0 mm Hg?

1.50 liters

17.6 liters

18.0 liters

34.2 liters

2. What is the pressure, in mm Hg, of 2.50 moles of an ideal gas if it has a volume of 50.0 liters when the temperature is 27.0° C?

84.2 mm Hg

289 mm Hg

617 mm Hg

936 mm Hg

3. When 12.4 grams of KBr are dissolved in enough water to create a 170-gram solution, what is the solution's concentration, expressed as a percent by mass?

6.2% KBr

6.8% KBr

7.3% KBr

7.9% KBr

1. Using the ideal gas law, which states that PV = nRT, and remembering that absolute temperature must be used:

(950.0 mmHg) (V) = (0.900 moles) (R) (25.0 + 273.15 K)

The most convenient value of R is 0.08206 L-atm/mol-K, so we convert the pressure of 950.0 mmHg to atm by dividing by 760: 1.25 atm

(1.25 atm) (V) = (0.9) (0.08206) (298.15)

V = 17.6 L

2. Again using the ideal gas law, and converting temperature to Kelvin: 27 + 273.15 = 300.15 K

PV = nRT

P (50.0 L) = (2.50 moles) (0.08206 L-atm/mol-K) (300.15 K)

P = 1.23 atm

Then we multiply by 760 mmHg / 1 atm = 936 mmHg. This is the fourth of the choices.

3. If the complete solution has a mass of 170 grams, and 12.4 grams of it is the dissolved KBr, we simply have to divide the mass of solute (KBr) by the total mass of the solution: This gives us 12.4 grams / 170 grams = 0.0729. Multiplying by 100% gives 7.29% or approximately 7.3% KBr, which is the third of the choices.