Solution.
P=P0(A)+P0(B)P = P^0(A) + P^0(B)P=P0(A)+P0(B)
P = 300 torr
χ(A)=P0(A)P=0.67\chi(A) = \frac{P^0(A)}{P} = 0.67χ(A)=PP0(A)=0.67
χ(B)=P0(B)P=0.33\chi(B) = \frac{P^0(B)}{P} =0.33χ(B)=PP0(B)=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=χ(a)×P0(A)+χ(B)×P0(B)
t=0.67×200+0.33×100=167 Kt = 0.67 \times 200 + 0.33 \times 100 = 167 \ Kt=0.67×200+0.33×100=167 K
50 is not a normal boiling point.
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