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 3 moles of A and 1 mole of B are added. (Given: Molecular mass of A and B are 40 and 60 respectively)

162.5 mm Hg

X_A = \(\frac{n_A}{n_A + n_B}\) = \(\frac{3}{3+1}\) = \(\frac{3}{4}\)

X_B = \(\frac{1}{4}\)

P_T = P_A + P_B = P^o_AX_A + P^o_BX_B = 200 x \(\frac{3}{4}\) + 50 x \(\frac{1}{4}\) = 162.5 mm Hg