MCQ
Benzoic acid undergoes dimerisation in benzene solution. The van't Hoff factor $(i)$ is related to the degree of association ' $x^{\prime}$ of the acid as
- A$i=(1-x)$
- B$i=(1+x)$
- C$i=\left(1-\frac{x}{2}\right)$
- ✓$i=\left(1+\frac{x}{2}\right)$
✓
Answer
Correct option: D.
$i=\left(1+\frac{x}{2}\right)$
d
$0.5\, \%$ solution of $KCl$
$0.5\, \%$ solution of $KCl$
So $m =\frac{0.5}{74.6} \times \frac{1}{0.1}$
$\Delta T _{ f }= i \times m _{ f }$
$0.24=i \cdot \frac{0.5}{74.6} \times \frac{1.80}{0.1}$
$i=\frac{0.24 \times 74.6}{0.5 \times 1.80} \times 0.1$
$=1.989$
$1.989=1+\alpha( n -1)$
$1.989=1+\alpha$
$\alpha=.989$
$\% \alpha=98.9 \%$
Ans $99 \,\%$
If mass of $H _{2} O =99.5$
$m =\frac{0.5}{74.5} \times \frac{1}{.0995}$
$i =\frac{0.24 \times 74.6 \times .0995}{.5 \times 1.80}$
$=1.979$
$1.979=1+\alpha( n -1)$
$1.979=1+\alpha$
$\alpha=.979$
$\% \alpha=97.9\, \%$
$98\, \%$
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