Question

Vapour Pressure of a mixture of benzene and toluene is given by $P=179X_B+92$, Where $X_B$ is mole fraction of benzene.
Vapour pressure of the solution obtained by mixing 936 g of benzene and 736 gm of toluene :

• A. 199.4 mm
• B. 271 mm
• C. 280 mm
• D. 289 mm

Answer 1

Answer: (A)

199.4 mm

P=179$X_B$+92

If [Benzene] =936g

If[Toluene]=736g

$X_B$= mole fraction of benzene in liquid

$X_B$=$\frac{\frac{936}{78}}{\frac{936}{78}+\frac{736}{92}}$=0.6

Now$P_B$=179$\times 0.6$ +92

$P_B$=199.4