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Q.
The amount of solute (molar mass $ 60\text{ }g\text{ }mo{{l}^{-1}} $ ) that must be added to $180\, g$ of water so that the vapour pressure of water is lowered by $10\%$, is
Relative lowering of vapour pressure is given by the formula
$ \frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{{{w}_{A}}}{{{M}_{A}}}\times \frac{{{M}_{B}}}{{{w}_{B}}} $
As vapour pressure of water is lowered by 10%.
$ \therefore $ $ \frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{10}{100} $
$ \therefore $ $ \frac{10}{100}=\frac{{{w}_{A}}}{60}\times \frac{18}{180} $
or $ {{w}_{A}}=60\text{ }g $