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

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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]

Ans.
(5)

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

   $i m k_{b} = 0.52 K$

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

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

   $i = 2$

${AB}_{2} ⇌ A^{2 +} + 2 B^{-}$

$i = 1 + (n - 1) α$

   $2 = 1 + (3 - 1) α$

$α = 0.5 = 5 \times 10^{- 1}$

Want Us to Email you About Special Offers & Updates?