At T(K) , the vapour pressure of the pure liquid A and B are 100mm and 160mm respectively. An ideal solution is formed 2 moles of A and 3 moles of B at the same temperature. The mole fraction of A and B in the vapour pressure respectively are
Answer (1)
Student Expert
25th Jun, 2021
Hi Aspirant!
As we know,
Vapour pressure of solution{P(total)} = P(A) + P(B), X(A)P(A) + X(B)P(B) {=>P(A)=X(A)p(A)}
The vapour pressure of component 1 , = P1=Y1P(total)
Here, y is the mole fraction of component 1 in vapur phase.
As, Vapour pressure of A (pure liquid) = P(A)= 100mm
and , Vapour pressure of B (pure liquid) = P(B)= 160mm
Therefore, the total vapour pressure of solution = P(A) + P(B)
P(Total) = X(A)P(A) + X(B)P(B) = (2/5)*100 + (3/5)*160 = 136mm
Also, P(A) = Y(A)P(total) {Y(A) is the mole fraction of A.}
Mole fraction of A = Y(A) = (P(A)/P(total)) = 40/136 = 0.294
Therefore, Y(B) = 1 - Y(A) = 1 - 0.294 = 0.706
Thankyou!
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