Solutions and Colligative Properties - Result Question 28-1
28. $75.2$ g of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$ (phenol) is dissolved in a solvent of $K_f=14$. If the depression in freezing point is $7$ K , then find the percentage of phenol that dimerises.
(2006, 2M)
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Answer:
Correct Answer: 28. $(75%)$
Solution:
$ \begin{aligned} & \text { Molar mass of solute }\left(M_B\right)=\dfrac{1000 \times K_f \times W_B}{W_A \times \Delta T_f} \\ & \Rightarrow \quad M_B=\dfrac{1000 \times 14 \times 75.2}{1000 \times 7} \\ & M_B=150.4 \mathrm{~g} \text { per mol } \\ & \end{aligned} $
Actual molar mass of phenol $=94 \mathrm{~g} / \mathrm{mol}$
Now, van’t Hoff factor, $i=\dfrac{\text { Calculated molar mass }}{\text { Observed molar mass }}$
$\therefore \quad i=\dfrac{94}{150.4}=0.625$
Dimerisation of phenol can be shown as :
Total number of moles at equilibrium, $i=1-\alpha+\dfrac{\alpha}{2}$
$ i=1-\dfrac{\alpha}{2} $
But $i=0.625$, thus,
$ \begin{aligned} 0.625 & =1-\dfrac{\alpha}{2} \\ \dfrac{\alpha}{2} & =1-0.625 \\ \alpha & =0.75 \end{aligned} $
Thus, the percentage of phenol that dimerises is $75 %$.
BETA
