An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanole are 2.619 kPa and 4.556 kPa, respectively, the composition of vapour (in terms of mole fraction) will be
A
0.635 MeOH, 0.365 EtOH
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B
0.365 MeOH, 0.635 EtOH
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C
0.574 MeOH, 0.326 EtOH
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D
0.173 MeOH, 0.827 EtOH
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Solution
The correct option is B 0.365 MeOH, 0.635 EtOH PEtOH=P∘× Mole fraction of EtOH in liquid phase =PT× Mole fraction of EtOH in vapour phase Thus, mole fraction of EtOH in vapour phase =P′PT=4.5564.556+2.619=0.635