Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at $100.15^{\circ} \mathrm{C}$. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at $-0.8^{\circ} \mathrm{C}$. The solutbility product of $\mathrm{PbCl}_{2}$ formed is ______ $\times 10^{-6}$ at 298 K . (Nearest integer) Given : $\mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ and $\mathrm{K}_{\mathrm{f}}=1.8 \mathrm{~kg} \mathrm{~mol}^{-1}$. Assume molality to be equal to molarity in all cases.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Solutions

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

100.1 5^{\circ} 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 solutbility product of

PbCl_{2}

formed 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.

Answer: 13

The boiling-point rise fixes the lead nitrate taken. Since

Pb (N O_{3})_{2}

gives three ions,

Δ T_{b} = 0.15 = K_{b} (3 a) = 0.5 \times 3 a

, so

a = 0.1 mol

, providing

0.1 mol P b^{2 +}

and

0.2 mol N O_{3}^{-}

. Adding

0.2 mol

sodium chloride precipitates

x

mol

PbC l_{2}

, leaving

P b^{2 +} = 0.1 - x

and

C l^{-} = 0.2 - 2 x

beside

0.2 mol

each of

N a^{+}

and

N O_{3}^{-}

. The freezing-point depression counts every dissolved particle,

0.8 = 1.8 [(0.1 - x) + (0.2 - 2 x) + 0.2 + 0.2] = 1.8 (0.7 - 3 x)

, giving

x = \frac{2.3}{27}

, so

[P b^{2 +}] = \frac{0.4}{27}

and

[C l^{-}] = \frac{0.8}{27}

. Then

K_{s p} = [P b^{2 +}] [C l^{-}]^{2} = \frac{0.4}{27} (\frac{0.8}{27})^{2} = 1.3 \times 1 0^{- 5}

, that is

13 \times 1 0^{- 6}

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer