Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15$^\circ$C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at $-0.8^\circ$ C. The solubility product of PbCl$_2$ formed is __________ $\times$ 10$^{-6}$ at 298 K. (Nearest integer)
Given : $\mathrm{K_b=0.5}$ K kg mol$^{-1}$ and $\mathrm{K_f=1.8}$ K kg mol$^{-1}$. Assume molality to the equal to molarity in all cases.
Answer (integer)
13
Solution
$0.15=3 \times 0.5 \times \mathrm{M}$
<br/><br/>
$$
\begin{aligned}
& \mathrm{M}_{\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}}=0.1 \text { molar } \\\\
& \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NaCl} \longrightarrow \mathrm{PbCl}_{2}+2 \mathrm{NaNO}_{3} \\\\
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \cdot \mathrm{m} \\\\
& 0.8=(0.4+3 s) 1.8 \\\\
& s=0.0148 \\\\
& \therefore \text { solubility product }=4 \mathrm{~s}^{3} \\\\
& =4 \times(0.0148)^{3} \\\\
& \approx 13 \times 10^{-6}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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