Question

3.795 g of sulphur is dissolved in 100 g of ${CS}_2$. This solution boils at 319.81 K. What is molecular formula of sulphur in solution? The boiling point of ${CS}_2$ is 319.45 K.
(Given that $K_b$ for ${CS}_2$ = 2.42 K kg ${mol}^{-1}$ and atomic mass of S = 32)

• A. $S_8$
• B. $S_4$
• C. $S_2$
• D. $S$

Answer 1

The correct option is A $S_8$

Let moles of sulphur be $n$ with no. of sulphur atoms in one molecule be $x$.

So, the molecule is of type $S_x$ ;

molecular mass will be $32\times x$

Molality (m) = $\frac{n}{100g}$ = $\frac{n}{0.1kg}$ = $10n$

$\because \Delta T_b=K_b\times m$

$⟹{T_b}^0-T_b=K_b\times m$

$⟹319.81-319.45=2.42\times 10n$

$⟹n=\frac{0.36}{24.2}$

$⟹\frac{3.795}{32\times x}=\frac{0.36}{24.2}$

$⟹x=8$

So, the molecule is $S_8$

Hence, Option "A" is the correct answer.