A one litre solution is prepared by dissolving some lead-nitrate in water. The solution was found to boil at `100.15^(@)C`. To the resulting solution `0.2` mole NaCI was added. The resulting solution was found to freeze at `0.83^(@)C`. Determine solubility product of `PbCI_(2)`. Given `K_(b) = 0.5` and `K_(f) = 1.86`. Assume molality to be equal to molarity in all case.

A one litre solution is prepared by dissolving some lead-nitrate in wa

1.	A one litre solution is prepared by dissolving some lead-nitrate in water. The solution was found to boil at `100.15^(@)C`. To the resulting solution `0.2` mole NaCI was added. The resulting solution was found to freeze at `0.83^(@)C`. Determine solubility product of `PbCI_(2)`. Given `K_(b) = 0.5` and `K_(f) = 1.86`. Assume molality to be equal to molarity in all case.
Answer» Correct Answer - `1.46 xx 10^(-5)` `Delta T_(b) = iK_(b)m` `0.15=3xx0.5xxm implies m=0.1` Now, `{:(Pb(NO_(3))_(2),+,2NaCl,rarr,PbCl_(2),+,2NaNO_(3)),(0.1,,0.2,,,,),(,,,,0.1,,0.2):}` Now, this solution constains two salts `Delta T_(f)=K_(f)xx m 0.83=1.86[2xx0.2+3s]` where s is molar solubility of `PbCl_(2)`. `s = 1.54xx10^(-3) K_(sp) = 4s^(3)=1.46xx10^(-5)`

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