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QJEEM23Ch9dbe: JEE Main 2023 Chemistry - Class 12 JEEM23Ch9dbe Solution: Solid Lead nitrate is dissolved in 1 litre of water. The…

Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15∘C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at −0.8∘ C. The solubility product of PbCl2 formed is __________ × 10−6 at 298 K. (Nearest integer)

Given : Kb=0.5 K kg mol−1 and Kf=1.8 K kg mol−1. Assume molality to the equal to molarity in all cases.

Step-by-Step Verified Solution

0.15=3×0.5×M0.15=3 \times 0.5 \times \mathrm{M}

MPb(NO3)2=0.1 molar Pb(NO3)2+2NaCl⟶PbCl2+2NaNO3ΔTf=iKf⋅m0.8=(0.4+3s)1.8s=0.0148∴ solubility product =4 s3=4×(0.0148)3≈13×10−6

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