Question

The vapour pressure of pure water at 25${}^o$C is 23.76 torr. The vapour pressure of a solution containing 5.40 g of a nonvolatile substance in 90.0 g water in 23.32 torr. Compute the molecular weight of the solute.

• A. $58.82$ g/mol
• B. $58.20$ g/mol
• C. $59.25$ g/mol
• D. $60.24$ g/mol

Answer 1

The correct option is A $58.82$ g/mol

The relationship between the relative lowering of vapour pressure and the molecular weight of solute and solvent is as shown below :

$\frac{\Delta P}{P}=\frac{W}{M}\times \frac{m}{w}$

Substitute values in the above expression :

$\frac{23.76-23.32}{23.76}=\frac{5.40}{M}\times \frac{18}{90.0}$

M = $58.82$ g/mol.

Hence, the molecular weight of solute is $58.82$ g/mol.

Therefore, option A is correct.