Question

1. Methane has a Henry's Law constant (k) of 9.8810–2 mol/(L·atm) when dissolved in benzene at 25°C. How many grams of CH4 (16.04 g/mole) will dissolve in 3.00 liters of benzene if the partial pressure of CH4 is 1.48 atm?

2. Carbon tetrachloride, once widely used in fire extinguishers and as a dry cleaning fluid, has been found to cause liver damage to those exposed to its vapors over long periods of time. What is the boiling point of a solution prepared by dissolving 375 g of sulfur (S8, 256.5 g/mol) in 1250 g of CCl4? Kb = 5.05°C/m, boiling point of pure CCl4 = 76.7°C?

3. What is the molarity of sodium (Na+) ion in a solution prepared by dissolving 0.100 moles of Na2SO4 and 0.050 moles of Na3PO4 in enough water to obtain 500.0 mL of solution.

Answer 1

Q1. Partial pressure CH₄ = 1.48 atm

Henry's law constant = 9.88 x 10^-2 mol/(L-atm)

Concentration of CH₄ in benzene = (Henry's law constant) * (Partial pressure CH₄)

Concentration of CH₄ in benzene = (9.88 x 10^-2 mol/(L-atm)) * (1.48 atm)

Concentration of CH₄ in benzene = 0.146 mol/L

moles CH₄ dissolved in benzene = (Concentration of CH₄ in benzene) * (volume of benzene)

moles CH₄ dissolved in benzene = (0.146 mol/L) * (3.00 L)

moles CH₄ dissolved in benzene = 0.439 mol

mass CH₄ dissolved in benzene = (moles CH₄ dissolved in benzene) * (molar mass CH₄)

mass CH₄ dissolved in benzene = (0.439 mol) * (16.04 g/mol)

mass CH₄ dissolved in benzene = 7.04 g

Q2. mass sulfur = 375 g

moles sulfur = (mass sulfur) / (molar mass S₈)

moles sulfur = (375 g) / (256.5 g/mol)

moles sulfur = 1.46 mol

molality sulfur = (moles sulfur) / (mass CCl₄ in kg)

molality sulfur = (1.46 mol) / (1.250 kg)

molality sulfur = 1.17 m

increase in boiling point = (K_b) * (molality sulfur)

increase in boiling point = (5.05 ^oC/m) * (1.17 m)

increase in boiling point = 5.91 ^oC

boiling point of solution = (boiling point of pure CCl₄) + (increase in boiling point)

boiling point of solution = (76.7 ^oC) + (5.91 ^oC)

boiling point of solution = 82.6 ^oC

Q3. moles Na₂SO₄ = 0.100 mol

moles Na⁺ = 2 * (moles Na₂SO₄)

moles Na⁺ = 2 * (0.100 mol)

moles Na⁺ = 0.200 mol

moles Na₃PO₄ = 0.050 mol

moles Na⁺ = 3 * (moles Na₃PO₄)

moles Na⁺ = 3 * (0.050 mol)

moles Na⁺ = 0.15 mol

Total moles Na⁺ = (0.200 mol) + (0.15 mol)

Total moles Na⁺ = 0.35 mol

Molarity Na⁺ = (Total moles Na⁺) / (volume of solution in Liter)

Molarity Na⁺ = (0.35 mol) / (0.500 L)

Molarity Na⁺ = 0.70 M