If the gas vapor mix is 0.2 mol A and 0.5 mole B where the total pressure is 40-mmHg, The partial pressure of A will NOT be 20-mmHg. It's more like 11.428-mmHg and B will be 28.572-mmHg.
0.2 mole A + 0.5 mole B => 0.7 mol mix.
mole fraction A = (0.2/0.7) = 0.2857
mole fraction B = (0.5/0.7) = 0.7143
∑mole fractions = 0.2857 + 0.7143 = 1.000
VP(A) = X₁∙VP₁ = 0.2857(40-mmHg) = 11.428-mmHg
VP(B) = X₂∙VP₂ = 0.7143(40-mmHg) = 28.572-mmHg
A mixture of 0.2mole of alcohol A and 0.5mole of alcohol has a total vapour pressure of 40mmHg at 298k,if the mixture obey's Raoult's law,find the pure pressure of B at 298k given that the pressure of A is 20mmHg at 298k
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