The vapour pressure of a solution of a non-volatile electrolyte B in a solvent A is 95% of the vapour pressure of the solvent at the same temperature. If the molecular weight of the solvent is 0.3 times the molecular weight of solute, the weight ratio of the solvent and solute are

The problem is based on Raoult's law, which states that the partial vapor 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.

Given that the vapor pressure of the solution is 95% of the vapor pressure of the solvent, we can write:

P_solution = 0.95 * P_solvent

According to Raoult's law, the vapor pressure of the solution is also equal to the mole fraction of the solvent (X_A) times the vapor pressure of the pure solvent:

P_solution = X_A * P_solvent

Setting these two equations equal to each other gives:

0.95 * P_solvent = X_A * P_solvent

Solving for X_A gives:

X_A = 0.95

The mole fraction of the solvent is also equal to the ratio of the number of moles of solvent to the total number of moles:

X_A = n_A / (n_A + n_B)

Given that the molecular weight of the solvent is 0.3 times the molecular weight of the solute, the number of moles of each can be expressed in terms of their weights (W_A and W_B) and molecular weights (M_A and M_B):

n_A = W_A / M_A n_B = W_B / M_B

Substituting these into the equation for X_A gives:

0.95 = (W_A / M_A) / ((W_A / M_A) + (W_B / M_B))

Solving this equation for the weight ratio of the solvent to the solute (W_A / W_B) gives:

W_A / W_B = 0.95 * (M_A / M_B) / (1 - 0.95)

Substituting M_A = 0.3 * M_B gives:

W_A / W_B = 0.95 * (0.3) / (1 - 0.95)

Solving this gives:

W_A / W_B = 0.285

So, the weight ratio of the solvent to the solute is approximately 0.285.