Question

a 35.0 gram sample of ethylene glycol, is dissolved in 500 grams of water The vapor pressure of pure water at 32.0 degrees Celsius at 35.7 torr. what is the vapor pressure of the solution at 32.0 degrees Celsius?

Answer 1

Mass of ethylene glycol , m₁ = 35 g

Mass of water , m₂ = 500g

Molar mass of etylene glycol , M₁ = 62.07g/mol

Molar mass of water, M₂ = 18.02g/mol

Step 1 : Calculating moles of ethylene glycol

moles of ethylene glycol , n₁ = Mass of ethylene glycol , m₁   Molar mass of etylene glycol ,M₁

moles of ethylene glycol , n₁ = 35g/(62.07 g/mol) = 0.564 mol

moles of ethylene glycol , n₁= 0.564 mol

Step 2 : Calculating moles of water

moles of water , n₂ = Mass of water, m₂ Molar mass of water,M₂

moles of water , n₂ = 500g / ( 18.02g/mol) = 27.75 mol

moles of water , n₂ = 27.75 mol

Step 3: calculating Mole fraction of water, x

Mole fraction of water, x = moles of water, n₂ Moles of water,n₂ + moles of ethylene glycol,n₁

Mole fraction of water, x = 27.75 / (27.75 + 0.564) = 0.98

Mole fraction of water = 0.98

Step 4: calculating vapor pressure of the solution

vapor pressure of pure water , P_water= 35.7 torr (given)

mole fraction of water, x_water = 0.98 (calculated in step 3)

Since, ethylene glycol is a non volatile compound , it will not contribute to the vapor pressure of the solution. Therefore, only water will contribute to the vapor pressure of the solution.

Using Raoult's law: P_solution = x_water P_water

P_solution = 0.98 X 35.7 torr

P_solution = 34.986 Torr

The vapor pressure of the solution at 32^oC = 34.986 Torr