Solid Lead nitrate is dissolved in 1litre of water. The solution was found boil at 100. 15°C. When 0.2 mol of NaCl is added to the resulting solution, it PbCl2 formed is ______ x 10-6 at 298K. (Nearest integer) Given: Kb = 0.5K kg mol-1 and kf 1.8K kg mol-1, Assume molality to be equal to molarity in all cases.

PbNO32⇒ΔTb=ikbm im=0.3 m=0.1NaCl=0.2mPbNO32+2NaCl→PbCl2+2NaNO3ΔTf=1.8i1m1+i2m20.8=1.8i1m1+2(0.2)818−0.4=i1m1i1m1=0.044m1=0.0443=0.0146= Solubility Ksp=4s3= =1.26×10−5 =12.6×10−6 ≅13×10−6

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Solid Lead nitrate is dissolved in 1litre of water. The solution was found boil at 100. 15°C. When 0.2 mol of NaCl is added to the resulting solution, it PbCl₂ formed is ______ x 10^-6 at 298K. (Nearest integer)
Given: K_b = 0.5K kg mol^-1 and kf 1.8K kg mol^-1, Assume molality to be equal to molarity in all cases.

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Detailed Solution

$\begin{array}{l} Pb {({NO}_{3})}_{2} \Rightarrow {ΔT}_{b} = {ik}_{b} m \\ im = 0.3 \\ m = 0.1 \\ NaCl = 0.2 m \\ Pb {({NO}_{3})}_{2} + 2 NaCl \to {PbCl}_{2} + 2 {NaNO}_{3} \end{array}$
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$\begin{array}{l} {ΔT}_{f} = 1.8 (i_{1} m_{1} + i_{2} m_{2}) \\ 0.8 = 1.8 (i_{1} m_{1} + 2 (0.2)) \\ \begin{aligned} (\frac{8}{18}) - 0.4 = i_{1} m_{1} \end{aligned} \\ \begin{aligned} i_{1} m_{1} = 0.044 \end{aligned} \\ \begin{aligned} m_{1} = \frac{0.044}{3} & = 0.0146 = Solubility \\ Ksp = 4 s^{3} = \\ = 1.26 \times 10^{- 5} \\ = 12.6 \times 10^{- 6} \\ ≅ 13 \times 10^{- 6} \end{aligned} \end{array}$