Two 5 molal solutions are prepared by dissolving a non-electroylete, non -volatile solute separately in the solvents X and Y. The molecular weights of the solvents are `M_(x)` and `M_(Y)`, respectively where `M_(X)=4/3M_(Y)`. The relative lowering of vapour pressure of the solution in X is "m" times that of the solution in Y. Given that the number of moles of solute is very small in comparision to that of solvent, the value of "m" is :

Options

A. `3/4`
B. `4/3`
C. `1/2`
D. `1/4`

Explanation

To solve the problem, we need to find the value of "m" which represents the relative lowering of vapor pressure of the solution in solvent X compared to that of solvent Y. Here’s the step-by-step solution:

### Step 1: Understand the given information
We have two 5 molal solutions prepared by dissolving a non-electrolyte, non-volatile solute in solvents X and Y. The molecular weights of the solvents are given as \( M_X \) and \( M_Y \), with the relationship \( M_X = \frac{4}{3} M_Y \).

### Step 2: Define relative lowering of vapor pressure
The relative lowering of vapor pressure (\( \Delta P/P \)) is given by the formula:
\[
...