The relative lowering of vapour pressure of an aqueous solution

 containing non-volatile solute is 0.0125. The molality of the solution is

Relative lowering of vapour pressure of an aqueous solution containing

 nonvolatile solute is equal to mole fraction of solute. 

$\begin{array}{l}\therefore \frac{P^{\circ }-P_s}{P^{\circ }}=\frac{n}{n+N}=0.0125\\ \Rightarrow \frac{n+N}{n}=\frac{1}{0.0125}\Rightarrow \frac{N}{n}=\frac{1}{0.0125}-1=\frac{0.9875}{0.0125}\\ \therefore \frac{N}{n}=\frac{0.9875}{0.0125}\end{array}$

Now, molality $=\frac{0.0125\times 1000}{0.9875\times 18}=0.70$