Two 5 molal solutions are prepared by dissolving a non-electrolyte, 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$$_X$$ = $$\frac{3}{4}$$ M$$_Y$$. 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 comparison to that of solvent, the value of "m" is:

We have two solutions that are both 5 molal. By definition, molality (usually denoted by $$b$$) is the number of moles of solute present in 1 kg (that is, 1000 g) of solvent:

$$b=\dfrac{n_{\text{solute}}}{\text{mass of solvent in kg}}$$

Given $$b=5\ \text{mol kg}^{-1}$$, the number of moles of solute present in exactly 1 kg of any solvent is

$$n_{\text{solute}} = 5\ \text{mol}$$

Now, for each solvent we need the number of moles of the solvent in that same 1 kg. If the molecular weight (molar mass) of a solvent is $$M$$, then

$$n_{\text{solvent}}=\dfrac{1000\ \text{g}}{M\ (\text{g mol}^{-1})}=\dfrac{1000}{M}\ \text{mol}$$

For solvent X the molar mass is $$M_X$$, so

$$n_{\text{solvent, X}}=\dfrac{1000}{M_X}$$

For solvent Y the molar mass is $$M_Y$$, so

$$n_{\text{solvent, Y}}=\dfrac{1000}{M_Y}$$

Next, recall the formula for the relative lowering of vapour pressure for a very dilute solution of a non-volatile, non-electrolyte solute:

$$\dfrac{p^{0}-p}{p^{0}}=X_{\text{solute}}$$

Here, $$X_{\text{solute}}$$ is the mole fraction of the solute. Because the solute is present in a very small amount compared with the solvent, we may write

$$X_{\text{solute}}=\dfrac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}\;\approx\;\dfrac{n_{\text{solute}}}{n_{\text{solvent}}}$$

Using the values already obtained:

For solvent X

$$X_{\text{solute, X}}\approx\dfrac{n_{\text{solute}}}{n_{\text{solvent, X}}} =\dfrac{5}{\dfrac{1000}{M_X}} =5\,\dfrac{M_X}{1000} =\dfrac{M_X}{200}$$

For solvent Y

$$X_{\text{solute, Y}}\approx\dfrac{n_{\text{solute}}}{n_{\text{solvent, Y}}} =\dfrac{5}{\dfrac{1000}{M_Y}} =5\,\dfrac{M_Y}{1000} =\dfrac{M_Y}{200}$$

The relative lowering of vapour pressure is numerically equal to these mole fractions, so the ratio of the two relative lowerings is

$$m=\dfrac{\left(\dfrac{p^{0}-p}{p^{0}}\right)_{\!X}} {\left(\dfrac{p^{0}-p}{p^{0}}\right)_{\!Y}} =\dfrac{X_{\text{solute, X}}}{X_{\text{solute, Y}}} =\dfrac{\dfrac{M_X}{200}}{\dfrac{M_Y}{200}} =\dfrac{M_X}{M_Y}$$

The problem states that the molar masses obey $$M_X=\dfrac{3}{4}\,M_Y.$$ Substituting this relation, we obtain

$$m=\dfrac{M_X}{M_Y}=\dfrac{3}{4}$$

Hence, the correct answer is Option A.