Question

What is the value of van’t Hoff factor (i) for $60\%$ association of $C{H}_{3}COOH$ in benzene?

• A. 0.5
• B. 0.4
• C. 0.7
• D. 0.9

Answer 1

The correct option is C 0.7

In benzene, two molecules of ethanoic acid associate to form a dimer.
$2C{H}_{3}COOH\rightleftharpoons (C{H}_{3}COOH{)}_2$
$Initiala0At.eqa(1-\alpha )a\frac{\alpha }{2}$
Number of moles at eq. $n_{eq}=a(1-\alpha +\alpha /2)$
At 60 % association $\alpha =0.6$
$\therefore n_{eq}=a(1-0.6+0.3)=a\left(0.7\right)$

van't Hoff factor i = $\frac{\text{number of solute particles in solution}}{\text{theoritical number of solute particles}}$
$i=\frac{a\left(0.7\right)}{a}=0.7$