The vapour pressure of benzene at a certain temperature is 640mm of Hg. A non volatile and non electrolyte solid weighing 2.175g is added to 39.08g of benzene. If the vapour pressure of the solution is 600mm of Hg, then the molecular weight of solid substance is

According to Raoult’s law, relative lowering of vapour pressure of solution is equal to mole fraction of the solute.

$\frac{\text{P}°-{\text{P}}_{\text{s}}}{\text{P}°}=\frac{{\text{w}}_1}{{\text{M}}_1}\times \frac{{\text{M}}_2}{{\text{W}}_2}\text{;}\text{eqn}\to \left(1\right)$

Here $\text{P}°$ is vapour pressure of pure solvent (benzene) $=640\text{mm}\text{of}\text{Hg}$

${\text{P}}_{\text{s}}$ is vapour pressure of solution $=600\text{mm}\text{of}\text{Hg}$

${\text{w}}_1=\text{weight}\text{of}\text{non}-\text{volatile}\text{solute}=2.175\text{g}$

${\text{W}}_2=\text{weight of solvent (benzene)}=39.08\text{g}$  

${\text{M}}_1=\text{molar mass of non-volatile solute}=?$

${\text{M}}_2=\text{Molar mass of solvent (benzene)}=78\text{g/mol}$

From equation 1,

$\frac{640-600}{640}=\frac{2.175}{{\text{M}}_1}\times \frac{78}{39.08}$

$\frac{40}{640}=\frac{2.175\times 2}{{\text{M}}_1}$

$\therefore {\text{M}}_1=\frac{2.175\times 2\times 640}{40}=69.60$