A solution containing 10 g of an electrolyte in 100 g of water boils at 100.52. The degree of ionization of the electrolyte is

Question

A solution containing 10 g of an electrolyte ${AB}_{2}$ in 100 g of water boils at 100.52 ${}^{°}C$ . The degree of ionization of the electrolyte ( $α$ ) is _____ $\times 10^{- 1}$ . (nearest integer)

[Given : Molar mass of ${AB}_{2} = 200 g$ ${mol}^{- 1}$ . $K_{b}$ (molal boiling point elevation const. of water) = 0.52 K kg ${mol}^{- 1}$ , boiling point of water = 100°C; ${AB}_{2}$ ionises as ${AB}_{2} \to A^{2 +} + 2 B^{-}$ ] [2024]

Smart Achievers · Accepted Answer

(5)

$Δ T_{b} = T_{b} - T_{b}^{°} = (100.52 - 100) K = 0.52 K$

$i m k_{b} = 0.52 K$

$i \frac{w_{solute} \times 1000}{M_{solute} \times W_{solvent} (in g)} k_{b} = 0.52 K$

$i \frac{10 \times 1000}{200 \times 100} \times 0.52 = 0.52$

$i = 2$

${AB}_{2} ⇌ A^{2 +} + 2 B^{-}$

$i = 1 + (n - 1) α$

$2 = 1 + (3 - 1) α$

$α = 0.5 = 5 \times 10^{- 1}$

Solution

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Solution

Ans. (5) ΔTb=Tb-Tb°=(100.52-100) K=0.52 K imkb=0.52 K iwsolute×1000Msolute×Wsolvent (in g)kb=0.52 K i10×1000200×100×0.52=0.52 i=2 AB2⇌A2++2B- i=1+(n-1)α 2=1+(3-1)α α=0.5=5×10-1

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