Question

The vapour pressure of pure benzene at ${50}^{o}C\text{is}268mm\text{of}Hg$. How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure $167mm\text{of}Hg$ at ${50}^{o}C$ ?

• A. $0.60$
• B. $0.25$
• C. $0.80$
• D. $0.32$

Answer 1

The correct option is A $0.60$

Applying Raoult's law in the following form:
$\frac{P_A^{\circ }-P_S}{P_S}=\frac{w_B\times M_A}{w_A\times M_B}=\frac{w_B/M_B}{w_A/M_A}$
$w_A=\text{mass of A (Solvent)}$
$w_B=\text{mass of B (Solute)}$
$M_A=\text{molar mass of A}$
$M_B=\text{molar mass of B}$
$\Rightarrow \frac{w_B/M_B}{w_A/M_A}$= Number of moles of solute per mole of benzene
$\Rightarrow \frac{n_B}{n_A}=\frac{P_A^{\circ }-P_S}{P_S}\Rightarrow \frac{n_B}{n_A}=\frac{(268-167)}{167}=0.6047\approx 0.605$