Solutions and Colligative Properties - Result Question 6-1
6. Molecules of benzoic acid $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right)$ dimerise in benzene. ’ $w$ ’ g of the acid dissolved in $30$ g of benzene shows a depression in freezing point equal to $2 K$ . If the percentage association of the acid to form dimer in the solution is $80$ , then $w$ is
(Given that $K_f=5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, molar mass of benzoic acid $=122 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(2019 Main, 12 Jan II)
(a) $1.8$ g
(b) $1.0$ g
(c) $2.4$ g
(d) $1.5$ g
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Answer:
Correct Answer: 6. (c)
Solution:
Molecules of benzoic acid dimerise in benzene as:
$ 2\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right) \rightleftharpoons\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right)_2 $
Now, we know that depression in freezing point $\left(\Delta T_f\right)$ is given by following equation:
$ \Delta T_f=i \times K_f \times m=\dfrac{i \times K_f \times w_{\text {solute }} \times 1000}{M w_{\text {solute }} \times w_{\text {solvent }}} \quad$ ……(i)
Given, $w_{\text {solute }}$ (benzoic acid) $=w g$
$ w_{\text {solvent }} \text { (benzene) }=30 \mathrm{~g} $
$ M w_{\text {Solute }} \text { (benzoic acid) }=122 \mathrm{~g} \mathrm{~mol}^{-1}, \Delta T_f=2 \mathrm{~K} $
$ K_f=5 K$ $ kg$ $ \mathrm{mol}^{-1}, % \alpha=80 \text { or } \alpha=0.8 $
Total number of moles at equilibrium $=0.2+0.4=0.6$
$ \begin{gathered} i=\dfrac{\text { Number of moles at equilibrium }}{\text { Number of moles present initially }} \\ i=\dfrac{0.6}{1}=0.6 \end{gathered} $
On substituting all the given values in Eq. (i), we get
$ 2=\dfrac{0.6 \times 5 \times w \times 1000}{122 \times 30}, \quad w=2.44 \mathrm{~g} $
Thus, weight of acid $(w)$ is $2.4$ g .
BETA
