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 -0.80 C. The solubility product of PbCl 2 formed is × 10-6 at 298 K. (Nearest integer) Given: K b =0.5 K kg mol -1 and K f =1.8 K kg mol -1. Assume molality to be equal to molarity in all cases.

Question

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 -0.80 C. The solubility product of PbCl 2 formed is × 10-6 at 298 K. (Nearest integer) Given: K b =0.5 K kg mol -1 and K f =1.8 K kg mol -1. Assume molality to be equal to molarity in all cases. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/6ac46ff5e361f619d83709ce469a67c8-.png" /> In final solution Δ T f =0.8=1.8[(0.3-3 x +0.2+0.2/1)] ⇒ x =(2.3/27) ⇒ K sp =(0.1-(2.3/27))(0.2-(4.6/27))2=13 × 10-6

In final solution ΔTf​=0.8=1.8[10.3−3x+0.2+0.2​] ⇒x=272.3​ ⇒Ksp​=(0.1−272.3​)(0.2−274.6​)2=13×10−6

In final solution
$Δ T_{f} = 0.8 = 1.8 [\frac{0.3 - 3 x + 0.2 + 0.2}{1}]$
$\Rightarrow x = \frac{2.3}{27}$
$\Rightarrow K_{s p} = (0.1 - \frac{2.3}{27}) (0.2 - \frac{4.6}{27})^{2} = 13 \times 1 0^{- 6}$