JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

Solid Lead nitrate is dissolved in $1$ litre of water. The solution was found to boil at $100.15^{\circ}C$ . When $0.2\mathrm{mol}$ of $\mathrm{NaCl}$ is added to the resulting solution, it was observed that the solution froze at $-0.8^{\circ}C$ . The solutbility product of ${\mathrm{PbCl}}_{2}$ formed is _____ $\times {10}^{-6}$ at $298K$ . (Nearest integer)

Given : ${K}_{b}=0.5K\mathrm{kg}{\mathrm{mol}}^{-1}$ and ${K}_{f}=1.8\mathrm{kg}{\mathrm{mol}}^{-1}$ . Assume molality to be equal to molarity in all cases.

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Held on 29 Jan 2023 · Verified 6 Jul 2026.

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Solution

Let a mole $\mathrm{Pb}{({\mathrm{NO}}_{3})}_{2}$ be added, dissociation of lead nitrate will result in:

$\begin{matrix}\mathrm{Pb}{({\mathrm{NO}}_{3})}_{2} & \rightarrow & {\mathrm{Pb}}^{2+} & + & 2{\mathrm{NO}}_{3}^{-} \\ a & & a & & 2a\end{matrix}$

$\begin{matrix}\Delta {T}_{b}=0.15= & 0.5[3a] & \Rightarrow & a=0.1 \\ & {\mathrm{Pb}}_{(\mathrm{aq})}^{2+} & + & 2{\mathrm{Cl}}_{(\mathrm{aq})}^{-}\rightarrow {\mathrm{PbCl}}_{2}(s)\end{matrix}$

During reformation of lead nitrate as a precipitate,

$\begin{matrix}{\mathrm{Pb}}_{(\mathrm{aq})}^{2+} & + & 2{\mathrm{Cl}}_{(\mathrm{aq})}^{-}\rightarrow {\mathrm{PbCl}}_{2}(s)\end{matrix}$

$\begin{matrix}t=0 & & 0.1 & & 0.2 \\ t=\infty & & 0.1 & & (0.2-2x)\end{matrix}$ ....(i)

In final solution after addition of 0.2 moles of sodium and chloride ions:

${\Delta T}_{f}={k}_{f}\times m \Rightarrow 0.8=1.8[\frac{0.3-3x+0.2+0.2}{1}]$

$\Rightarrow x=\frac{2.3}{27}$

From (i),

$\Rightarrow {K}_{\mathrm{sp}}=[{\mathrm{Pb}}^{2+}][2{\mathrm{Cl}}^{-}{]}^{2} \Rightarrow (0.1-\frac{2.3}{27}){(0.2-\frac{4.6}{27})}^{2}=13\times {10}^{-6}$

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Solution

Let a mole $\mathrm{Pb}{({\mathrm{NO}}_{3})}_{2}$ be added, dissociation of lead nitrate will result in:

$\begin{matrix}\mathrm{Pb}{({\mathrm{NO}}_{3})}_{2} & \rightarrow & {\mathrm{Pb}}^{2+} & + & 2{\mathrm{NO}}_{3}^{-} \\ a & & a & & 2a\end{matrix}$

$\begin{matrix}\Delta {T}_{b}=0.15= & 0.5[3a] & \Rightarrow & a=0.1 \\ & {\mathrm{Pb}}_{(\mathrm{aq})}^{2+} & + & 2{\mathrm{Cl}}_{(\mathrm{aq})}^{-}\rightarrow {\mathrm{PbCl}}_{2}(s)\end{matrix}$

During reformation of lead nitrate as a precipitate,

$\begin{matrix}{\mathrm{Pb}}_{(\mathrm{aq})}^{2+} & + & 2{\mathrm{Cl}}_{(\mathrm{aq})}^{-}\rightarrow {\mathrm{PbCl}}_{2}(s)\end{matrix}$

$\begin{matrix}t=0 & & 0.1 & & 0.2 \\ t=\infty & & 0.1 & & (0.2-2x)\end{matrix}$ ....(i)

In final solution after addition of 0.2 moles of sodium and chloride ions:

${\Delta T}_{f}={k}_{f}\times m \Rightarrow 0.8=1.8[\frac{0.3-3x+0.2+0.2}{1}]$

$\Rightarrow x=\frac{2.3}{27}$

From (i),

$\Rightarrow {K}_{\mathrm{sp}}=[{\mathrm{Pb}}^{2+}][2{\mathrm{Cl}}^{-}{]}^{2} \Rightarrow (0.1-\frac{2.3}{27}){(0.2-\frac{4.6}{27})}^{2}=13\times {10}^{-6}$

JEE Main 2023 — Chemistry Physical Chemistry

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JEE Main 2023 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry