Question

Calculate the vapor pressure of a solution containing 27.0 g of glycerin (C3H8O3) in 119 mL of water at 30.0 ∘C. The vapor pressure of pure water at this temperature is 31.8 torr. Assume that glycerin is not volatile and dissolves molecularly (i.e., it is not ionic) and use a density of 1.00 g/mL for the water.

Answer 1

Answer – Given, mass of glycerin (C₃H₈O₃) = 27.0 g , volume of water = 119 mL

Density of water = 1.0 g/mL, at temp 30^oC vapor pressure of pure water = 31.8 torr

We know,

Mass = density * volume

          = 1.0 g/mL * 119 mL

          = 119 g

So, moles of glycerin (C₃H₈O₃) = 27.0 g / 92.09 g.mol^-1 = 0.293 moles

moles of water = 119 g / 18.015 g.mol^-1 = 6.605 moles

mole fraction of water = moles of water / total moles

                                = 6.605 / 0.293 + 6.605

                                = 0.957

We know the formula

Vapor pressure of solution = mole fraction of water * Pure pressure of solvent

                                           = 0.957 * 31.8 torr

                                          = 30.4 torr