Question

A solution is prepared by dissolving 10 gram of non volatile solute in 200 grams of water. It has a vapour pressure of 31.84 mmhg at 308 k. Calculate the molar mass of the solute( vapour pressure of pure water at 308 k is equal to 32 mm hg

Answer 1

please give me answer

Answer 2

Answer:

The molar mass of the solute is 180.90 g/mol.

Explanation:

Vapor pressure of the solution =$p_s= 31.84 mmHg$

Vapor pressure of water = $p_o=32 mmHg$

Moles of solvent =$n_1=\frac{200 g}{18 g}=11.11 mol$

Moles of non volatile solute = $\frac{10}{m}$

The relative lowering of the vapor pressure of solution is equal to the mole fraction of solute:

$\frac{(p^o-p^s)}{p^o}=\frac{n_2}{n_1+n_2}$

$\frac{32 mmHg-31.84 mmHg}{32 mmHg}=\frac{\frac{10 g}{m}}{11.11 mol+\frac{10 g}{m} mol}$

m = 180.90 g/mol

The molar mass of the solute is 180.90 g/mol.