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(A) Let $a$ moles of $Pb(NO_3)_2$ be added.$Pb(NO_3)_2 \rightarrow Pb^{2+} + 2NO_3^-$Initial moles: $a \quad 0 \quad 0$Final moles: $0 \quad a \quad 2

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) Let $a$ moles of $Pb(NO_3)_2$ be added.$Pb(NO_3)_2 ightarrow Pb^{2+} + 2NO_3^-$Initial moles: $a \quad 0 \quad 0$Final moles: $0 \quad a \quad 2a$Total particles = $3a$.$\Delta T_b = i 	imes K_b 	imes m = 3 	imes 0.5 	imes a = 0.15 \Rightarrow a = 0.1 \, mol$.Now,$Pb^{2+} + 2Cl^- ightarrow PbCl_2(s)$.Initial moles: $Pb^{2+} = 0.1, Cl^- = 0.2, NO_3^- = 0.2$.Let $x$ moles of $PbCl_2$ precipitate.Remaining moles: $Pb^{2+} = (0.1-x), Cl^- = (0.2-2x), NO_3^- = 0.2$.Total moles of solute particles = $(0.1-x) + (0.2-2x) + 0.2 = 0.5-3x$.$\Delta T_f = K_f 	imes m = 1.8 	imes (0.5-3x) = 0.8$.$0.9 - 5.4x = 0.8$ $\Rightarrow 5.4x = 0.1$ $\Rightarrow x = 0.1/5.4 = 1/54$.$[Pb^{2+}] = 0.1 - 1/54 = 4.4/54 \approx 0.0815$.$[Cl^-] = 0.2 - 2/54 = 8.8/54 \approx 0.163$.$K_{sp} = [Pb^{2+}][Cl^-]^2 = (4.4/54) 	imes (8.8/54)^2 \approx 13 	imes 10^{-6}$.

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