Liquid A and B form an ideal solution at 30^oC. At this temperature vapour pressure of pure A and pure B is 200 mm Hg and 50 mm Hg respectively. Calculate the vapour pressure of the solution at same temperature if equal weight of A and B are added. (Given: Molecular mass of A and B are 40 and 60 respectively)

140 mm Hg

n_A = \(\frac{x}{40}\), n_B = \(\frac{x}{60}\)

X_A = \(\frac{n_A}{n_A + n_B}\) 

X_A = \(\frac{\frac{x}{40}}{\frac{x}{40} + \frac{x}{60}}\) = 0.6

and X_B = 1 - 0.6 = 0.4

P_T = P_A + P_B = P^o_AX_A + P^o_BX_B = 200 x 0.6 + 50 x 0.4 = 120 + 20 = 140 mm Hg