Q60 / 180 · JIPMER 2015 · Chemistry
The molal freezing point depression constant for benzene $\left(\mathrm{C}_6 \mathrm{H}_6\right)$ is $4.90 \mathrm{~K} \mathrm{~kg} \mathrm{mol}^{-1}$. Selenium exists as a polymer of the type $\mathrm{Se}_x$. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is $0.112^{\circ} \mathrm{C}$ lower than that of pure benzene. The molecular formula of selenium is (atomic mass of $\mathrm{Se}=78.8 \mathrm{~g} \mathrm{~mol}^{-1}$)
- Correct answerA$\mathrm{Se}_8$
- B$\mathrm{Se}_6$
- C$\mathrm{Se}_4$
- D$\mathrm{Se}_2$
Solution
To calculate the molar mass of selenium (Se)
$M_B=\frac{K_f \times W_B}{\Delta T_f \times W_A}$
Mas of selenium $\left(W_B\right)=3.26 \mathrm{~g}$
Mass of benzene $\left(W_A\right)=226 \mathrm{~g}=0.226 \mathrm{~kg}$
Depression in freezing point $\left(\Delta T_f\right)=0.112^{\circ} \mathrm{C}$
Molar depression constant $\left(K_f\right)=4.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
$\therefore M_B=\frac{\left(4.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)(3.26 \mathrm{~g})}{(0.112 \mathrm{~K})(0.226 \mathrm{~kg})}=631.08 \mathrm{~g} \mathrm{~mol}^{-1}$
Given, gram atomic mass of selenium
$=78.8 \mathrm{~g} / \mathrm{mol}$
For $\mathrm{Se}_x\left(78.8 \mathrm{~g} \mathrm{~mol}^{-1}\right)=631.08 \mathrm{~g} \mathrm{~mol}^{-1}$
or, $\quad x=\frac{631.08 \mathrm{~g} \mathrm{~mol}^{-1}}{78.8 \mathrm{~g} \mathrm{~mol}^{-1}}=8$
Thus, molecular formula of selenium $=\mathrm{Se}_8$
Hence, the correct answer is option: 'A'.