Read the Passage and answer the following question:

There are many properties of ideal solutions which depends only on the concentration of the solute particles and are independent of the nature of the solute. Such properties are called colligative properties. For example, when a non-volatile solute is added to a solvent, its vapour pressure gets lowered. Raoult established that relative lowering in vapour pressure depends only on the concentration of the solute and it is independent of its identity. Similarly, elevation in boiling point, depression in freezing point and osmotic pressure on addition of a non volatile solute to a solvent depend only on the concentration of solute but are independent of its nature. These properties can also be used to determine the molar mass of the solute. The relative lowering in vapour pressure changes when the solute is electrolyte and undergoes association or dissociation in the solution. The molar mass obtained in these electrolytic solutions are abnormal molar masses. 

When 3.6 g of a non volatile solute 'X' is dissolved in 90 g of water at 293 K, its vapour pressure is lowered by 0.4%.What is the Relative lowering in vapour pressure (Vapour pressure of water at 293 K is 17.5 mm Hg)?

0.004

The correct answer is option 3. 0.004.

To find the relative lowering in vapor pressure, we need to understand the concept of relative lowering of vapor pressure, which is given by:

\(\frac{\Delta P}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} \)

where

\(\Delta P\) is the lowering of vapor pressure,

\(P_0\) is the vapor pressure of the pure solvent,

\(n_{\text{solute}}\) is the number of moles of the solute,

\(n_{\text{solvent}}\) is the number of moles of the solvent.

Given:

The vapor pressure of pure water (\(P_0\)) at 293 K = 17.5 mm Hg,

The vapor pressure is lowered by 0.4%,

Mass of solute \(X\) = 3.6 g

Mass of water = 90 g,

Molar mass of water = 18 g/mol.

Lowering of vapor pressure (\(\Delta P\)) = 0.4% of 17.5 mm Hg,

\(\Delta P = \frac{0.4}{100} \times 17.5 = 0.07 \text{ mm Hg}\).

\(\frac{\Delta P}{P_0} = \frac{0.07}{17.5} = 0.004\).

Therefore, the relative lowering in vapor pressure is 3. 0.004.