Solutions and Colligative Properties - Result Question 13-1
13. The freezing point of benzene decreases by $0.45^{\circ} \mathrm{C}$ when $0.2$ g of acetic acid is added to $20$ g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be ( $K_f$ for benzene $=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
(2017 Main)
(a) $64.6 %$
(b) $80.4 %$
(c) $74.6 %$
(d) $94.6 %$
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Answer:
Correct Answer: 13. (d)
Solution:
Let the degree of association of acetic acid $\left(\mathrm{CH}_3 \mathrm{COOH}\right)$ in benzene is $\alpha$, then
$\therefore$ Total moles $=1-\alpha+\dfrac{\alpha}{2}=1-\dfrac{\alpha}{2}$ or $i=1-\dfrac{\alpha}{2}$
Now, depression in freezing point $\left(\Delta T_f\right)$ is given as
$ \Delta T_f=i K_f m \quad$ …….(i)
where, $K_f=$ molal depression constant or cryoscopic constant.
$ \begin{gathered} m=\text { molality } \\ \text { Molality }=\dfrac{\text { number of moles of solute }}{\text { weight of solvent (in kg) }}=\dfrac{0.2}{60} \times \dfrac{1000}{20} \end{gathered} $
Putting the values in Eq. (i)
$\therefore 0.45 =[1-\dfrac{\alpha}{2}](5.12)[\dfrac{0.2}{60} \times \dfrac{1000}{20}] $
$1-\dfrac{\alpha}{2} =\dfrac{0.45 \times 60 \times 20}{5.12 \times 0.2 \times 1000} $
$\therefore 1-\dfrac{\alpha}{2} =0.527 \Rightarrow \dfrac{\alpha}{2}=1-0.527 $
$\alpha =0.946$
Thus, percentage of association $=94.6 %$
BETA
