Question

The vapour pressure of $C_6{H}_6$ and $C_7{H}_8$ mixture at ${50}^{\circ }C$ is given by $p=179X_B+92$ where $X_B$ is the mole fraction of $C_6{H}_6$.

Calculate (in mm) Vapour pressure of liquid mixture obtained by mixing $936g{C}_6{H}_6$ and $736g$ toluene is:

• A. 300 mm Hg
• B. 250 mm Hg
• C. 199.4 mm Hg
• D. 180.6 mm Hg

Answer 1

The correct option is C 199.4 mm Hg

Given,PM = 179 $X_B$+92

For pure $C_6{h}_6$, $x_B,$ i.e. mole fraction of benzene $=1$

$\Rightarrow P_B^0=179+92=271mm$

$x_B=\frac{\frac{936}{78}}{\frac{936}{78}+\frac{736}{92}}=\frac{12}{12+8}=0.60$

$\begin{array}{cc}{x}_T& =\frac{\frac{736}{92}}{\frac{736}{92}+\frac{936}{78}}=\frac{8}{12+8}=0.40\\ P_M=& 179x_B+92\\ =& 179\times 0.6+92\\ =& 107.4+92\\ =& 199.4mm\text{of}Hg\\ \text{Hence,}\left(C\right)\text{is correct option.}& \end{array}$