Question #116439
PA°equal to 200 torr and PB°equals to 100 torr calculate the composition of vapour of liquid solution containing A and B at 50 degree centigrade is the normal boiling point of solution?
1
Expert's answer
2020-05-19T08:43:39-0400

Solution.

P=P0(A)+P0(B)P = P^0(A) + P^0(B)

P = 300 torr

χ(A)=P0(A)P=0.67\chi(A) = \frac{P^0(A)}{P} = 0.67

χ(B)=P0(B)P=0.33\chi(B) = \frac{P^0(B)}{P} =0.33

t=χ(a)×P0(A)+χ(B)×P0(B)t = \chi(a) \times P^0(A) + \chi(B) \times P^0(B)

t=0.67×200+0.33×100=167 Kt = 0.67 \times 200 + 0.33 \times 100 = 167 \ K

50 is not a normal boiling point.

Answer:

χ(A)=P0(A)P=0.67\chi(A) = \frac{P^0(A)}{P} = 0.67

χ(B)=P0(B)P=0.33\chi(B) = \frac{P^0(B)}{P} =0.33

50 is not a normal boiling point.


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