QJEEAdv22Ch514c: JEE Advanced 2022 Chemistry - Class 12 JEEAdv22Ch514c Solution: An aqueous solution is prepared by dissolving of an ionic salt…

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 .