JEE MAIN - Chemistry (2023 - 29th January Morning Shift - No. 17)
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
13
Explanation
$0.15=3 \times 0.5 \times \mathrm{M}$
$$ \begin{aligned} & \mathrm{M}_{\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}}=0.1 \text { molar } \\\\ & \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NaCl} \longrightarrow \mathrm{PbCl}_{2}+2 \mathrm{NaNO}_{3} \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \cdot \mathrm{m} \\\\ & 0.8=(0.4+3 s) 1.8 \\\\ & s=0.0148 \\\\ & \therefore \text { solubility product }=4 \mathrm{~s}^{3} \\\\ & =4 \times(0.0148)^{3} \\\\ & \approx 13 \times 10^{-6} \end{aligned} $$
$$ \begin{aligned} & \mathrm{M}_{\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}}=0.1 \text { molar } \\\\ & \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NaCl} \longrightarrow \mathrm{PbCl}_{2}+2 \mathrm{NaNO}_{3} \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \cdot \mathrm{m} \\\\ & 0.8=(0.4+3 s) 1.8 \\\\ & s=0.0148 \\\\ & \therefore \text { solubility product }=4 \mathrm{~s}^{3} \\\\ & =4 \times(0.0148)^{3} \\\\ & \approx 13 \times 10^{-6} \end{aligned} $$
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