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At 20°C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20°C above an equimolar mixture of benzene and methyl benzene is ________ . (Nearest integer)
 

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Given,

\begin{aligned} &P_{\text {Benzene }}^{0}=P_{B}^{0}=70 \text { torr } \\ &P_{\text {Tolvene }}^{0}=P_{T}^{0}=20 \text { torr } \end{aligned}

\chi_{B}=0.5, \chi_{T}=0.5

\begin{aligned} \therefore P_{\text {Total }} &=P_{B}^{0} \cdot \chi_{B}+P_{T} \cdot \chi_{T} \\ &=70 \times \frac{1}{2}+20 \times \frac{1}{2} \\ &=35+10=45 \text { torr } \end{aligned}

And , \mathrm{P_{B}=P_{B}^{0} \chi_{B}=70 \times \frac{1}{2}=35 {torr}}

\begin{aligned} \therefore \quad \text{Mole fraction of Benzene in Vapor} (y_{B})=\frac{35}{45} &=0.78 \\ &=78 \times 10^{-2} \end{aligned}

Hence, the correct answer is 78.

Posted by

sudhir.kumar

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