A solution containing 10g of an electrolyte $\mathrm{AB}_{2}$ in 100 g of water boils at $100.52^{\circ} \mathrm{C}$. The degree of ionization of the electrolyte ($\alpha$) is ______ $\times 10^{-1}$. (nearest integer) [Given : Molar mass of $\mathrm{AB}_{2}=200 \mathrm{~g} \mathrm{~mol}^{-1}$. $\mathrm{K}_{\mathrm{b}}$ (molal boiling point elevation const. of water) $=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, boiling point of water $=100^{\circ} \mathrm{C}$; $\mathrm{AB}_{2}$ ionises as $\mathrm{AB}_{2} \rightarrow \mathrm{~A}^{2+}+2 \mathrm{~B}^{-}$]

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Solutions

A solution containing 10g of an electrolyte

AB_{2}

in 100 g of water boils at

100.5 2^{\circ} C

. The degree of ionization of the electrolyte (

α

) is ______

\times 1 0^{- 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

= 10 0^{\circ} C

;

AB_{2}

ionises as

AB_{2} \to A^{2 +} + 2 B^{-}

]

Answer: 5

Solution

Δ T_{b} = i K_{b} m

. Molality

= \frac{10/200}{100/1000} = 0.5

m. So

0.52 = i (0.52) (0.5) \Rightarrow i = 2

. For

A B_{2} \to A^{2 +} + 2 B^{-}

i = 1 + 2 α \Rightarrow 2 = 1 + 2 α \Rightarrow α = 0.5 = 5 \times 1 0^{- 1}

, so the answer is 5.

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Integer answer

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