Benzoic acid undergoes dimerization in benzene. $x \mathrm{~g}$ of benzoic acid (molar mass $122 \mathrm{~g}…

Benzoic acid undergoes dimerization in benzene. $x \mathrm{~g}$ of benzoic acid (molar mass $122 \mathrm{~g} \mathrm{~mol}^{-1}$) is dissolved in 49 g of benzene. The depression in freezing point is 1.12 K. If degree of association of acid is $88 \%$, what is the value of $x$ ? $\left(\mathrm{~K}_{\mathrm{f}}\right.$ for benzene $\left.=4.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
  1. 2.44
  2. 1.22
  3. 3.66
  4. 4.88

Solution

Given, weight of benzoic acid $=\mathrm{xg}$
Molar mass of benzoic acid $=122 \mathrm{~g} \mathrm{~mol}^{-1}$ Weight of benzene $=49 \mathrm{~g}$ Molar mass of benzene $=78 \mathrm{~g} \mathrm{~mol}^{-1}$
$\begin{aligned} & \text { depression in freezing point }\left(\Delta \mathrm{T}_{\mathrm{f}}\right)=1.12 \mathrm{k} \\ & \text { degree of association }(\alpha)=88 \% \\ & =0.88 \\ & \begin{aligned} \mathrm{K}_{\mathrm{F}} \text { for benzene }=4.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\end{aligned} \\ & \text { For association, } \\ & \text { Van't Hoff factor }(\mathrm{i})=1+\left(\frac{1}{\mathrm{n}}-1\right) \alpha \\ & =1+\left(\frac{1}{2}-1\right) \times 0.88 \\ & \begin{aligned} \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}\end{aligned} \\ & \begin{aligned} 1.12=0.56\end{aligned}\end{aligned}$ $1.12=0.56 \times 4.9 \times \frac{\mathrm{x} \times 1000}{122 \times 49}\left[\mathrm{~m}=\frac{\text { weight of solute } \times 100}{\begin{array}{l}\text { Molecular } \times \text { Weight } \\ \text { Weight of of solvent } \\ \text { solute }\end{array}}\right]$ $\mathrm{x}=2.44 \mathrm{~g}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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