Question

A solution containing 28 g of phosphorus in $315\,g\,CS_{2}$ (b.p. $46.3^{\circ}C)$ boils at $47.98^{\circ}C$ If $K_{b}$ for $CS_{2}$ is $2.34\,K\,Kg\,mol^{-1}$ The formula of phosphorus is (at. mass of P = 31)

• A. $P_{6}$
• B. $P_{4}$
• C. $P_{3}$
• D. $P_{2}$

Answer 1

Answer: (B)

$\Delta T_{b}=m\,K_{b}=\frac{w}{M} \times\frac{1000}{W}\times K_{b}$
$\Delta T_{b}=47.98-46.3=1.68$
$1.68=\frac{28}{M}\times\frac{1000}{315}\times2.38$
$M=\frac{28\times1000\times2.38}{315\times1.68}=125.92$
Atomicity $= \frac{\text{Mol.wt.}}{\text{At.wt.}}=\frac{125.92}{31}=4.02$
So. Molecule is $=P_{4}$