Question

The molal freezing point depression constant for benzene $\left(C_6{H}_6\right)$ is $4.90KKgmo{l}^{-1}$. Selenium exists as a polymer of the type $S{e}_x$, of benzene, the observed freezing point is ${0.112}^{\circ }C$ lower than that of pure benzene. The molecular formula of selenium is:
(Atomic mass of $Se=78.8gmo{l}^{-1})$

• A. $S{e}_8$
• B. $S{e}_6$
• C. $S{e}_4$
• D. $S{e}_2$

Answer 1

The correct option is A $S{e}_8$

To calculate the molar mass of selenium ($Se$)

$M_B=\frac{K_f\times W_B}{\Delta T_f\times W_A}$

Mass of selenium $\left(W_B\right)=3.26$ g

Mass of benzene $\left(W_A\right)=226$ g $=0.226$ kg

Depression in freezing point $\left(\Delta T_f\right)=0.112{}^{o}C$

$\therefore M_B=\frac{4.9\times 3,26}{0.112\times 0.226}=631.08$ g mol${}^{-1}$

Given, gram atomic mass of selenium $=78.8$ g/mol

For $S{e}_x=631.08$ g mole${}^{-1}$

or, $x=\frac{631.08}{78.8}=8$