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

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 $- 0. 8^{\circ} C$ . The solubility product of $PbCl_{2} (K_{sp} = [Pb^{2 +}] [Cl^{-}]^{2})$ is $____$ $\times 1 0^{- 6}$ at 298 K . (Nearest integer)

Given : K_b = 0.5 K kg mol^-1 and K_f = 1.8 kg mol^-1 . Assume molality to be equal to molarity in all cases.

TestHub · Accepted Answer

Let a mole of Pb ( N O 3 ​ ) 2 ​ be added. Pb ( N O 3 ​ ) 2 ​ → Pb 2 + + 2 N O 3 ​ − a a 2 a Δ T b ​ = 0.15 = 0.5 [ 3 a ] ⇒ a = 0.1 Pb 2 + ( aq ) + 2 Cl − ( aq ) → PbCl 2 ​ ( s ) t = 0 0.1 0.2 t = ∞ ( 0.1 − x ) ( 0.2 − 2 x ) In the final solution: Δ T f ​ ​ = 0.8 = 1.8 [ 1 0.3 − 3 x + 0.2 + 0.2 ​ ] ⇒ x = 27 2.3 ​ ⇒ K s p ​ = ( 0.1 − 27 2.3 ​ ) ( 0.2 − 27 4.6 ​ ) 2 = 13 × 1 0 − 6 ​

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