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JEE MAIN 2023
29-1-23
Question
Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15ºC. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at –0.8º C. The solutbility product of PbCl2 formed is ______ × 10–6 at 298 K. (Nearest integer) Given : Kb = 0.5 K kg mol–1 and Kf = 1.8 kg mol–1. Assume molality to be equal to molarity in all cases.
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Solution
Sol. Let a mole $\mathrm{Pb}\left(\mathrm{NO}_3\right)_2$ be added $$ $$ In final solution $$ \begin{array}{ll} & \Delta T_f=0.8=1.8\left[\frac{0.3-3 x+0.2+0.2}{1}\right] \\ \Rightarrow & x=\frac{2.3}{27} \\ \Rightarrow & K_{x p}=\left(0.1-\frac{2.3}{27}\right)\left(0.2-\frac{4.6}{27}\right)^2=13 \times 10^{-6} \end{array} $$
Question Tags
JEE Main
Chemistry
Easy
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