Question

what is the vapor pressure of a solution containing 0.500 mol of glucose and 5.000 mol of water at 29 degrees C?

Answer 1

To solve this, we need vapor pressure of pure water at 29 °C

At 29°C, vapor pressure of pure water = 30.0 torr

Number of moles of glucose ( a non-volatile solute) = 0.500 mol

Number of moles of water = 5.000 mol

Total number of moles = 0.500 + 5.000 = 5.500 mol

Mole fraction of water = 5.000/5.500 = 0.909091

According to Raoult’s law,

when a non-volatile solute is dissolved in solvent, the vapor pressure of the solution depends only on the mole fraction of the pure solvent.

So,

vapor pressure of the solution = mole fraction of water * vapor pressure of water

Vapor pressure of the solution = 0.909091 * 30.0 torr = 27.3 torr