Two liquids having vapor pressure $P_1^o$ and $P_2^o$ in pure state in the ratio of $2:1$ are mixed in the molar ratio of $1:2$. The ratio of their moles in the vapor state would be :

According to Raoult's Law, the mole fraction in the vapor phase y_i is given by y_i = (P_i^o * x_i) / P_total. Given P1^o = 2k, P2^o = k, and molar ratio x1:x2 = 1:2, then x1 = 1/3 and x2 = 2/3. The partial pressures are p1 = 2k * 1/3 = 2k/3 and p2 = k * 2/3 = 2k/3. Since partial pressures are equal, the mole ratio in vapor is 1:1.