Question

A solution containing 28 g phosphorous in 315 g $C{S}_2$ (b.pt. ${46.3}^{o}C$) boils at ${47.98}^{o}C,K_b$ for $C{S}_2$ is 2.34 K $mo{l}^{-1}kg$. Its molecular formula is $P_x$. Find $x$. Assume its complete association.

Answer 1

$\Delta T=47.98-46.3=1.68$

$\Delta T=\frac{1000\times K_f\times w}{m\times W}$

$\therefore 1.68=\frac{1000\times 2.3\times 28}{m\times 315}$

$\therefore m_{exp}=123.80$

The molar mass of phosphorus is 123.80 g/mol.
$\because$ For association, $m_N/m_{exp}=1-\alpha +\left(\alpha /n\right)$
$\therefore \frac{31}{123.80}=1-1+\frac{1}{n}$

$[\because m_N$ of $P=31$ and $\alpha =1$ (given)]
$\therefore n=4$, i.e., molecule exists as $P_4$.