Question

$1$ mole benzene $\left(p_{\text {benzene }}^{o}=42\, mm \right)$ and $2$ mole toluene $\left(p_{\text {benzene }}^{n}=36\, mm \right)$ will have:

• A. Mole fraction of benzene in vapour phase above liq. Mixture is $1 / 3$.
• B. Mole fraction of toluene in vapour phase above liq. Mixture is $8 / 19$
• C. Total vapour pressure of solution is $38 mm$
• D. Mole fraction of benzene in vapour phase above liquid mixture is $7 / 19$

Answer 1

Answer: (C), (D)

$P_{\text {total }}=P_{B}^{o} X_{B}+P_{T}^{o} X_{T}$
$=42 \times \frac{1}{3}+36 \times \frac{3}{2}=38 \,mm$
$P _{ M } Y _{ B }= P _{ B }^{ o } \times X _{ B }$ where $YB$ is the mole fraction of Benzene is vapour phase
$38 \times Y_{B}=14 ; Y_{B}=\frac{14}{38}=\frac{7}{19}$
$Y _{ A }=1-\frac{7}{19}=\frac{11}{19}$