Question

The vapour pressure of pure benzene at a certain temperature is $640mmHg$. A non - valatile solid weighing $2.175g$ is added to $39.0g$ of benzene. The Vapour pressure of the solution is $600mm$ Hg. What is the molecular mass of the solid substance?

• A. $65.25g$
• B. $40.25g$
• C. $25.25g$
• D. $80.25g$

Answer 1

The correct option is A $65.25g$

According to Raoult's law,
$\frac{P_A^o-P_s}{P_A^o}=\frac{n_B}{n_B+n_A}$
$n_B=\text{number of moles of non volatile solid}{n}_A=\text{number of moles of benzene}$

Let m be the molecular mass of the solid substance.
$n_B=\frac{2.175}{m};n_A=\frac{39}{78}=0.5\text{(Molecular mass of benzene = 78)}{P}_o=640mm;P_s=600mm$
Substituting the value in the above equation,
$\frac{640-600}{640}=\frac{\frac{2.175}{m}}{\frac{2.175}{m}+0.5}=\frac{2.175}{2.175+0.5m}m=\frac{2.175\times 16-2.175}{0.5}=65.25g$