Question

liquid A(p°A=360 mm hg) and B ( p° B = 320 mm hg) are mixed. if solution has vapour pressure 340 mm hg. then mole fraction of B in the solution would be?? Please answer this question ​

Answer 1

hey mate!

Vapour Pressure of A, poA = 360 mm Hg

Vapour Pressure of B, poB = 320 mm Hg

Vapour Pressure of Solution, P = 340 mm Hg

We know,

P = χApoA + χB poB

Because, Sum of the mole fractions, i.e., χA + χB = 1

Therefore, the equation becomes,

P = (1 - χB) poA + χB poB

P = poA + ( poB - poA) χB

Placing the values, we get -

340 mm Hg = 360 mm Hg + (320 - 360)mm Hg χB

340 mm Hg =360 mm Hg - 40 mm Hg χB

(320 - 360) mm Hg = - 40 mm Hg χB

So, 40 mm Hg χB = 20 mm Hg

Thus, χB = 2040 = 12

Answer 2

Answer:

Xb=1/2

Explanation:

Ptotal =Pa*(1-Xb) + Pb*Xb

340 = 360 (1-Xb) + 320*Xb

20=40*Xb

Xb =1/2