Calculate the vapor pressure of a solution containing 27.2 g of glycerin (c3h8o3) in 132 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.

This problem is to apply Roult's Law.

Roult's Law states that the vapor pressure, p, of a solution of a non-volatile solute is equal to the vapor pressure of the pure solvent, Po solv, times the mole fraction of the solvent, Xsolv

p = Xsolv * Po sol

X solv = number of moles of solvent / number of moles of solution

The solvent is water and the solute (not volatile) is glycerin.

Number of moles = mass in grams / molar mass

mass of water = 132 ml * 1 g/ml = 132 g

molar mass of water = 18 g/mol

=> number of moles of water = 132 g / 18 g/mol = 7.33333 mol

mass of glycerin = 27.2 g

molar mass of glycerin:, C3H8O3: 3 * 12 g/mol + 8 * 1 g/mol + 3*16 g/mol = 92 g/mol

number of moles of glycerin = 27.2g / 92 g/mol = 0.29565

total number of moles = 7.33333 moles + 0.29565 moles = 7.62898 moles

=> X solv = 7.33333 / 7.62898 = 0.96125

=> p = 0.96125 * 31.8 torr ≈ 30.57 torr ≈ 30.6 torr.

Answer: 30.6 torr