Realtive lowering in vapour pressure of a solution containing `1` mole `K_(2)SO_(4)` in `54 g H_(2)O` is (`K_(2)SO_(4)` in `100%` ionised) :

Options

A. `(1)/(55)`
B. `(3)/(55)`
C. `(3)/(4)`
D. `(1)/(2)`

Explanation

To solve the problem of calculating the relative lowering in vapor pressure of a solution containing 1 mole of K₂SO₄ in 54 g of H₂O, we can follow these steps:

### Step 1: Determine the number of moles of the solvent (H₂O).
The molecular weight of water (H₂O) is approximately 18 g/mol.

\[ \text{Moles of H₂O} = \frac{\text{mass of H₂O}}{\text{molar mass of H₂O}} = \frac{54 \text{ g}}{18 \text{ g/mol}} = 3 \text{ moles} \]

### Step 2: Determine the total number of moles in the solution.
We have 1 mole of K₂SO₄, which dissociates into 3 ions (2 K⁺ and 1 SO₄²⁻) when it fully ionizes. Therefore, the total number of moles of solute (ions) is:

\[ \text{Total moles of K₂SO₄} = 1 \text{ mole} \quad \text{(which gives 3 moles of ions)} \]

Thus, the total number of moles in the solution is:

\[ \text{Total moles} = \text{moles of solute} + \text{moles of solvent} = 3 + 1 = 4 \text{ moles} \]

### Step 3: Calculate the mole fraction of the solute.
The mole fraction (X) of the solute (K₂SO₄) is given by:

\[ X_{\text{solute}} = \frac{\text{moles of solute}}{\text{total moles}} = \frac{1}{4} \]

### Step 4: Calculate the van 't Hoff factor (i).
Since K₂SO₄ dissociates into 3 ions, the van 't Hoff factor (i) is:

\[ i = 3 \]

### Step 5: Calculate the relative lowering of vapor pressure.
The relative lowering of vapor pressure (ΔP) can be calculated using Raoult's law:

\[ \frac{P_0 - P_s}{P_0} = X_{\text{solute}} \cdot i \]

Substituting the values we have:

\[ \frac{P_0 - P_s}{P_0} = \left(\frac{1}{4}\right) \cdot 3 = \frac{3}{4} \]

### Final Result
Thus, the relative lowering in vapor pressure of the solution is:

\[ \frac{P_0 - P_s}{P_0} = \frac{3}{4} \]