The vapor pressure of a naqueous solution is found to be 24.9 mmgh at 25C. what is the mole fraction of solute in this solution?

Mole fraction of solute is 0.0462

Explanation:

To solve this we use the colligative property of lowering vapor pressure.

First of all, we search for vapor pressure of pure water at 25°C  = 23.8 Torr

Now, we convert the Torr to mmHg. Ratio is 1:1, so 23.8 Torr is 23.8 mmHg.

Formula for lowering vapor pressure is:

ΔP = P° . Xm

Where ΔP = P' (Vapor pressure of solution) - P° (Vapor pressure of pure solvent)

Xm = mole fraction

24.9 mmHg - 23.8 mmHg = 23mmHg . Xm

Xm = (24.9 mmHg - 23.8 mmHg) /  23mmHg

Xm = 0.0462

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa