Question

Benzoic acid undergoes dimerisation in benzene solution. The van 't Hoff factor $\left(i\right)$ is related to the degree of association ${}^{\prime }{x}^{\prime }$ of the acid as:

• A. $i=(1-x)$
• B. $i=(1+x)$
• C. $i=(1-x/2)$
• D. $i=(1+x/2)$

Answer 1

The correct option is C $i=(1-x/2)$

$\begin{array}{cccc}& 2C_6{H}_{5}COOH& \rightleftharpoons & (C_6{H}_{5}COOH{)}_2\\ \text{Before}& 1& & 0\\ \text{After}& 1-x& & x/2\end{array}$
$\therefore \text{Total}=1-x+x/2=1-x/2$
$\therefore i=\frac{1-x/2}{1}=1-x/2$