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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

$AB_2 $ is 10% dissociated in water to $A^{2+}$ and $B^-$ ....

$AB_2 $ is 10% dissociated in water to $A^{2+}$ and $B^-$. The boiling point of a 10.0 molal aqueous solution of $AB_2 $ is _ _ _ _ $^ \circ C $. (Round off to the nearest integer)

[Given: Molal elevation constant of water $K_b  = 0.5 K.Kg.mol^{-1} $, boiling point of pure water = $100 ^\circ C $]

Solution

At ionic equilibrium, $\mathop {A{B_2}}\limits_{a(1 - \alpha )} \rightleftharpoons \mathop {{A^{2 + }}}\limits_{a\alpha }  + \mathop {2{B^ - }}\limits_{2a\alpha } $

Total moles of particles $=a(1-\alpha) + a\alpha + 2a\alpha = a(1+2\alpha ) = 1.2a $

$\therefore i=\frac {1.2a}{a} = 1.2 $ 

$\Delta T_b = i K_b m = 1.2 \times 0.5 \times 10 = 6 $

$\therefore B.P. = 106^\circ C $