Question

The values of Van’t Hoff factors for KCl, NaCl and K2SO4, respectively, are
_____________.
(i) 2, 2 and 2
(ii) 2, 2 and 3
(iii) 1, 1 and 2
(iv) 1, 1 and 1

NCERT Class XII
Chemistry - Exemplar Problems

Chapter 2. Solutions

Answer 1

(ii) 2 ,2 and 3  -  is the answer.

Answer 2

Answer : The correct option is, (ii) 2, 2 and 3

Explanation :

Van't Hoff factor is the ratio of number of particles after association to the number of particles before association.

$i=\frac{\text {observed colligative property}}{\text {Calculated Colligative property}}$

For 100% dissociation of a solute, van't Hoff factor is equal to the number of ions produces from one molecule of the solute.

For KCl :

$KCl\rightarrow K^++Cl^-$

After dissociation : the number of ions produced = 1 + 1 = 2

So, the Van't Hoff factor, $i=2$

For NaCl :

$NaCl\rightarrow Na^++Cl^-$

After dissociation : the number of ions produced = 1 + 1 = 2

So, the Van't Hoff factor, $i=2$

For $K_2SO_4$ :

$K_2SO_4\rightarrow 2K^++SO_4^{2-}$

After dissociation : the number of ions produced = 2 + 1 = 3

So, the Van't Hoff factor, $i=3$

Therefore, the correct option is, (ii) 2, 2 and 3