Question

Binary salt $\stackrel{+}{\text{A}}\stackrel{-}{\text{B}}$ 40% dissociated and 60% associated. Van't Hoff factor for the solution will be:

• A. 1.1
• B. 1.4
• C. 1.5
• D. 2

Answer 1

The correct option is A 1.1

Solution:- (A) $1.1$

Van't Hoff factor is given as-

$i=\frac{n}{n_0}$

Whereas,

$n$ is the no. of particles after dissolving

$n_0$ is the initial no. of particles

According to the question-

$i=\frac{n_{diss.}+n_{ass.}}{n_0}$

Whereas

$n_{diss.}$ is the number of particles appeared after dissociation of $40\%$ of molecules $AB$:

$AB\rightleftharpoons A^{+}+B^{-}$

$n_{ass.}$ is the number of particles appeared after association of $60\%$ of molecules AB:

$2AB\rightleftharpoons {\left(AB\right)}_2$

$\therefore i=\frac{2n_0\times 0.4+\frac{n_0}{2}\times 0.6}{n_0}=0.8+03=1.1$

Hence the Van't Hoff factor for the solution is $1.1$.