Question

The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass $78gmo{l}^{-1})$. The vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance?

• A. $170gmo{l}^{-1}$
• B. $270gmo{l}^{-1}$
• C. $370gmo{l}^{-1}$
• D. $140gmo{l}^{-1}$

Answer 1

The correct option is A $170gmo{l}^{-1}$

The given values are $P^{\circ }=0.850bar,Ps=0.845bar;M{w}_1=78gmo{l}^{-1}$ $W_2=0.5g;W_1=39g$ Now using equation
$\frac{P^{\circ }-P_s}{P^{\circ }}=\frac{W_1\times M{w}_1}{M{w}_2\times W_1}$ On substituting all the given values, we get $\frac{0.850-0.845}{0.850}=\frac{0.5g\times 78gmo{l}^{-1}}{M{w}_2\times 39g}$ $\therefore M{w}_2=170gmo{l}^{-1}$