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The Vapour pressure of pure benzene is 70 torr and that of methylbenzene is 20 torr. If we have an equimolar mixture of the above two components, find the mole fraction of benzene in the vapour phase.

Solution: Answer:7/9 \(\begin{array}{l}P_{C_{6}H_{6}}^{0} = 70\: torr\end{array} \) \(\begin{array}{l}P_{C_{6}H_{5}CH_{3}}^{0} = 20\: torr\end{array} \) \(\begin{array}{l}X_{C_{6}H_{6}} = X_{C_{6}H_{5}CH_{3}}\end{array} \) \(\begin{array}{l}Y_{C_{6}H_{6}} = \frac{P_{C_{6}H_{6}}^{0} \times... View Article