Question

Dry air was passed through bulbs containing a solution of $40\text{g}$ of non-volatile electrolytic solute in $360\text{g}$ of water, then through bulbs containing pure water at the same temperature and finally through a tube in which pumice moistened with strong $H_{2}S{O}_4$ was kept. The water bulbs lost $0.0870\text{g}$ and the sulphuric acid tube gained $2.175\text{g}$. Calculate the molar mass of solute (in gram)

Answer 1

By using the formula,
$\frac{P-P_S}{P}=x_{solute}$
$x=\text{Mole fraction of solute}=\frac{\text{Number of moles of solute}}{\text{Total number of moles}}$

Substituting the values, we get
$\frac{0.087}{2.175}=\frac{\frac{40}{M}}{\frac{40}{M}+\frac{360}{18}}$
$\Rightarrow 25=1+\frac{M}{2}$
$\Rightarrow M=48\text{g}$