A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of $$A = 100\,\text{g mol}^{-1}$$; $$B = 200\,\text{g mol}^{-1}$$; $$C = 10,000\,\text{g mol}^{-1}$$]

We have to compare the relative lowering of vapour pressure produced when the same mass (10 g) of three different non-volatile solutes A, B and C is dissolved in the same mass (180 g) of water.

For a dilute solution containing a non-volatile solute, Raoult’s law gives

$$P = P^{\circ}X_{\text{solvent}}$$

where $$P^{\circ}$$ is the vapour pressure of the pure solvent and $$X_{\text{solvent}}$$ is its mole fraction. Re-arranging, the relative lowering of vapour pressure is

$$\frac{P^{\circ}-P}{P^{\circ}} = 1 - X_{\text{solvent}} = X_{\text{solute}}.$$

That is,

$$\frac{\Delta P}{P^{\circ}} = X_{\text{solute}}.$$

The mole fraction of the solute equals

$$X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}.$$

Because the solutions are very dilute ($$n_{\text{solute}} \ll n_{\text{solvent}}$$), we can write approximately

$$X_{\text{solute}} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}.$$

So, for a fixed quantity of solvent, the relative lowering is directly proportional to the number of moles of solute present. We therefore calculate the moles of each solute.

The moles of water (solvent) are first obtained. The molar mass of water is 18 g mol-1, hence

$$n_{\text{solvent}} = \frac{180\ \text{g}}{18\ \text{g mol}^{-1}} = 10\ \text{mol}.$$

Now, for each solute:

1. Solute A: molar mass $$M_A = 100\ \text{g mol}^{-1}$$.    Moles of A:

$$n_A = \frac{10\ \text{g}}{100\ \text{g mol}^{-1}} = 0.1\ \text{mol}.$$

2. Solute B: molar mass $$M_B = 200\ \text{g mol}^{-1}$$.    Moles of B:

$$n_B = \frac{10\ \text{g}}{200\ \text{g mol}^{-1}} = 0.05\ \text{mol}.$$

3. Solute C: molar mass $$M_C = 10\,000\ \text{g mol}^{-1}$$.    Moles of C:

$$n_C = \frac{10\ \text{g}}{10\,000\ \text{g mol}^{-1}} = 0.001\ \text{mol}.$$

Next, we obtain the approximate relative lowering for each case using $$X_{\text{solute}} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}.$$

For A:

$$X_A \approx \frac{0.1}{10} = 0.01.$$

For B:

$$X_B \approx \frac{0.05}{10} = 0.005.$$

For C:

$$X_C \approx \frac{0.001}{10} = 0.0001.$$

We can now arrange these values in decreasing order:

$$0.01 \; (A) \; > \; 0.005 \; (B) \; > \; 0.0001 \; (C).$$

Therefore, the relative lowering of vapour pressure follows the order

$$A > B > C.$$

This corresponds to Option C (the third option in the list).

Hence, the correct answer is Option C.