The vapour pressure of $C_6H_6$ and $C_7H_8$ mixture at $50^{\circ}C$ is given by $p=179X_B+92$ where $X_B$ is the mole fraction of $C_6H_6$. Calculate (in mm) Vapour pressure of liquid mixture obtained by mixing $936 \, g\ C_6H_6 \, $ and $736\ g$ toluene is:

First calculate the moles of benzene (C6H6) by dividing 936 g by its molar mass of 78 g/mol to get 12 moles, and the moles of toluene by dividing 736 g by its molar mass of 92 g/mol to get 8 moles. The mole fraction of benzene, X_B, is 12 divided by the total moles (12 + 8 = 20), yielding 0.6. Substituting X_B = 0.6 into the given vapour pressure equation gives p = 179(0.6) + 92, which equals 107.4 + 92 = 199.4 mm Hg. The vapour pressure of the liquid mixture is 199.4 mm Hg.

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