Equal amount of solute are dissloved in equal amount of two solvents A and B. The realative lowering of vapour pressure for the solution of A has twice the relative lowering of vapour pressure for solution of B. If M_A and M_B are molar mass of solvent A and B respectively, then for dilute solution of A and B with solute-

A) M_A=M_B

B)M_A=M_B/2

C)M_A=4M_B

D)M_A=2M_B

The relative lowering of vapour pressure is directly proportional to mole fraction of solute in solution.
Suppose the relative lowering of vapor pressure of solution with solvent A is P_A.
The relative lowering of vapor pressure of solution with solvent B is P_B
The mole fraction of solute in solvent A is X_A
​Mole fraction of solute in solution B is X_B
​P_A = kX_A
P_B = kX_B

​Given that, P_A = 2P_B
​So, kX_A = 2KX_B
​Or, X_A = 2 X_B             ........(1)

X_A =$\frac{{\text{W}}_{\text{solute}}{\text{/M}}_{\text{solute}}}{{\text{W}}_{\text{solute}}{\text{/M}}_{\text{solute}}{\text{+W}}_{\text{solvent A}}{\text{/M}}_{\text{solvent A}}}$

X_B = $\frac{{\text{W}}_{\text{solute}}{\text{/M}}_{\text{solute}}}{{\text{W}}_{\text{solute}}{\text{/M}}_{\text{solute}}{\text{+W}}_{\text{solvent B}}{\text{/M}}_{\text{solvent B}}}$

Let the weight of solute and molar mass is 1​
X_A = M_A/W_A
X_B = M_B/W_B
Putting the value of X_A and X_B in equation 1 we get
 M_A/W_A = 2 M_B/W_B 
Given that W_A is equal to W_B
therefore M_A = 2M_B
Option D is correct.