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  • Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at . W

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 0C. The solubility product ofPbCl_{2} formed is_____\times 10^{-6}at 298 K. (Nearest integer) 

(Given : \mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K} \mathrm{kgmol}^{-1} and \mathrm{K}_{\mathrm{f}}=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}  Assume molality to be equal to molarity in all cases.)

Option: 1

13


Option: 2

"__"


Option: 3

"__"


Option: 4

"__"


Answers (1)

best_answer

Let a mole \mathrm{Pb}\left(\mathrm{NO}_3\right)_2  be added

\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{Pb}^{2+}+2 \mathrm{NO}_3^{-}

\begin{aligned} & \text { a } \quad \text { a } \quad 2 a \\ & \Delta \mathrm{T}_{\mathrm{b}}=0.15=0.5[3 \mathrm{a}] \Rightarrow \mathrm{a}=0.1 \\ & \mathrm{~Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cl}_{(\mathrm{aq})}^{-} \rightarrow \mathrm{PbCl}_2(\mathrm{~s}) \\ & \end{aligned}

\begin{array}{lll}\mathrm{t}=0 & 0.1 & 0.2 \\ \mathrm{t}=\infty & (0.1-\mathrm{x}) & (0.2-2 \mathrm{x})\end{array}

In final solution

\begin{aligned} & \Delta \mathrm{T}_{\mathrm{f}}=0.8=1.8\left[\frac{0.3+3 \mathrm{x}+0.2+0.2}{1}\right] \\ & \Rightarrow \mathrm{x}=\frac{2.3}{27} \\ & \Rightarrow \mathrm{K}_{\mathrm{sp}}=\left(0.1-\frac{2.3}{27}\right)\left(0.2-\frac{4.6}{27}\right)^2=13 \times 10^{-6}\end{aligned}

Posted by

Ritika Kankaria

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