Question

4.) Pure water has a vapor pressure of 31.1 torr at 30.1 °C. A solution is prepared by adding 90.8 g of unknown substance which is a nonvolatile and non-electrolyte to 375.0 g of water. The vapor pressure of the resulting solution is 29.5 torr. Calculate the molar mass of the unknown substance.

Answer 1

Solution

Given data

• Amount of solute m_s = 90.8 g
• Amount of solvent m_water = 375 g
• Vapor pressure of pure solvent P_solvent = 31.1 torr
• Vapor pressure of solution P_solution = 29.5 torr
• Molar mass of water M_water = 18 g / mol
• Molar mass of Solute = M_s

We know that,

29.5 = X_water * 31.1

X_water = 0.9485

X_water is mole fraction of water in solution.

$X_{water} = \frac{N_{water}}{N_{water}+N_{s}}$

N = Number of moles

N_water = m_water / M_water

= 375 / 18 = 20.833 mole

From above equation,

$0.9485 = \frac{20.833}{20.833+N_{s}}$

N_s = 1.1311 mole

Now we calculate molar mass.

N_s = m_s / M_s

1.1311 = 90.8 / M_s

M_s = 80.2758 g / mol

Answer

• Molar mass of solute = 80.2758 g / mol