100 g of liquid ‘A’ (molar mass 140 gmol-1)was dissolved in 1000 g of liquid ‘B’ (molar mass 180 gmol-1).The vapour pressure of liquid ‘B’ was found to be 500 torr. Calculate the vapour pressure of pureliquid ‘A’ if the total vapor pressure of the solution is 475 torr.

The problem can be solved using Raoult's law, which states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

First, we need to calculate the mole fraction of each component in the solution.

The number of moles of liquid A is given by mass/molar mass = 100 g / 140 g/mol = 0.714 mol. The number of moles of liquid B is given by mass/molar mass = 1000 g / 180 g/mol = 5.56 mol.

The mole fraction of A, X_A, is given by the number of moles of A divided by the total number of moles: X_A = 0.714 / (0.714 + 5.56) = 0.114. The mole fraction of B, X_B, is given by the number of moles of B divided by the total number of moles: X_B = 5.56 / (0.714 + 5.56) = 0.886.

According to Raoult's law, the partial pressure of A, P_A, is given by the vapor pressure of the pure component, P_A^0, multiplied by its mole fraction: P_A = P_A^0 * X_A. Similarly, the partial pressure of B, P_B, is given by the vapor pressure of the pure component, P_B^0, multiplied by its mole fraction: P_B = P_B^0 * X_B.

The total vapor pressure of the solution, P_total, is given by the sum of the partial pressures of A and B: P_total = P_A + P_B.

We know that P_B = 500 torr, X_B = 0.886, and P_total = 475 torr. We can substitute these values into the equation for P_total to find P_A:

475 torr = P_A + 500 torr * 0.886 P_A = 475 torr - 500 torr * 0.886 = 43.0 torr.

Finally, we can substitute the values of P_A and X_A into the equation for P_A to find P_A^0:

43.0 torr = P_A^0 * 0.114 P_A^0 = 43.0 torr / 0.114 = 377 torr.

So, the vapor pressure of pure liquid A is 377 torr.