A current of dry air was passed through a solution of $2.5$ g of a non-volatile solute in $100$ g of water and through water alone. The loss in the weight of solution was $1.25$ g and that of water was $0.005$ g. Calculate the molecular mass of the solute.

$P_o-P_S\propto$ loss in weight of pure water

and $P_S\propto$ loss in weight of solution

Using $\frac{P_o-P_S}{P_s}=\frac{n_2}{n_1}=\frac{m_2\times M_1}{M_2\times m_1}$

Given that:

loss of weigt os solution=1.25 g and that of pure water = 0.005g

mass of solute, $m_2=2.5g$

mass of solvent (water), $m_1=100g$

Molar mass of solute $M_2$

molar mass of water, $M_1=18g/mol$

substituting the values in above equation we get:

$\frac{0.005}{1.25}=\frac{2.5\times 18}{M_2\times 100}$

$M_2=112.5g/mol$

molar mass of solute is 112.5 g/mol