Question

An aqueous sodium sulfide solution has a vapor pressure of 80.9mmHg at 50 ?C. The vapor pressure of pure water at this temperature is 92.6 mmHg. (Assume sodium sulfide completely dissociates.)

What is the concentration of sodium sulfide in mass percent?

Please show all work.

Answer 1

Vapor pressure of solution = 80.9 mmHg

vapor pressure of pure water = 92.6 mmHg

mole fraction of water = (Vapor pressure of solution) / (vapor pressure of pure water)

mole fraction of water = (80.9 mmHg) / (92.6 mmHg)

mole fraction of water = 0.874

let moles of water = 1 mol

let moles of sodium sulfide = x

Total moles = (moles of water) + (moles of sodium sulfide)

Total moles = 1 mol + x

mole fraction of water = (moles of water) / (Total moles)

0.874 = (1 mol) / (1 mol + x)

1 mol + x = (1 mol) / (0.874)

1 mol + x = 1.145 mol

x = 1.145 mol - 1 mol

x = 0.145 mol

moles sodium sulfide = 0.145 mol

mass sodium sulfide = (moles sodium sulfide) * (molar mass sodium sulfide)

mass sodium sulfide = (0.145 mol) * (78.0452 g/mol)

mass sodium sulfide = 11.3 g

mass water = (moles water) * (molar mass water)

mass water = (1 mol) * (18.0 g/mol)

mass water = 18.0 g

Total mass = (mass sodium sulfide) + (mass water)

Total mass = (11.3 g) + (18.0 g)

Total mass = 29.3 g

mass percent sodium sulfide = (mass sodium sulfide / total mass) * 100

mass percent sodium sulfide = (11.3 g / 29.3 g) * 100

mass percent sodium sulfide = 38.6 %