Vapour pressure of solution $\left(\mathrm{P}_{\mathrm{A}}\right)$ $$ =59.724 \mathrm{~mm} \text { of } \mathrm{Hg} $$

Vapour pressure of pure water $\left(\mathrm{P}_{\mathrm{A}}^{\circ}\right)$ $$ =60.000 \mathrm{~mm} \text { of } \mathrm{Hg} $$

Also, $0.1 \mathrm{~mol}$ of an ionic solid is dissolved in $1.8 \mathrm{~kg}$ of water and salt remains $90 \%$ dissocated in the solution.



So, total number of moles $=0.01+0.09 a$ of non-volatile particles.

Now, mass of water $=1.8 \mathrm{~kg}=1.8 \times 1.8 \times 1000 \mathrm{~g}$

Molar mass of water $=18 \mathrm{~g}$

Moles of water $=\frac{1.8 \times 1000}{18}=100$ moles

Using the colligative property, relative lowering in vapour pressure,

$$ \frac{\mathrm{P}_{\mathrm{A}}^{\circ}-\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}^{\circ}}=x_{\mathrm{A}} $$

$$ \Rightarrow $$ $\frac{60-59.724}{60} =\frac{0.01+0.09 a}{100} $

$$ \Rightarrow $$ $ \frac{0.276}{60} =\frac{0.01+0.09 a}{100} $

$$ \Rightarrow $$ $ \frac{27.6}{60} =0.01+0.09 a $

$$ \Rightarrow $$ $ 0.46 =0.01+0.09 a $

$$ \Rightarrow $$ $ 0.09 a =0.45 $

$$ \Rightarrow $$ $ a =\frac{0.45}{0.89} $

$$ \Rightarrow $$ $ a =5 $

So, the number of ions present per formula unit of the ionic salt is 5 .