Read the passage carefully and answer the questions.

The osmotic pressure is one of the four colligative properties which depends on the number of particles of solute in the solution, irrespective of their nature. The osmotic pressure is equal to the product of concentration, gas constant and temperature. It is the most commonly used property for the determination of the molar mass of biomolecules. In many situations, the molecules get dissociated or associated when they are dissolved in a solvent, thereby changing the number of molecules in the solution. As a result, an abnormal molecular mass is obtained. The extent of dissociation or association is given by van't Hoff factor (i), which is the ratio of normal molar mass to abnormal molar mass. It is related with the degree of association or ionization. The introduction of van't Hoff factor in the osmotic pressure equation modifies the equation.

The osmotic pressure of a 0.1 M aqueous solution of $K_2SO_4$ is 4.92 atm at 300 K. The percentage ionization of the salt is:

(Given $R = 0.082\, atm\, L\, K^{-1}\, mol^{-1}$)

50

The correct answer is Option (1) → 50

Given:

• $C = 0.1 \, \mathrm{M}$
• $П_\text{observed} = 4.92 \, \mathrm{atm}$
• $T = 300 \, \mathrm{K}$
• $R = 0.082 \, \mathrm{L·atm·K^{-1}·mol^{-1}}$
• Salt: ${K_2SO_4}$ → dissociates into 2 K⁺ + 1 SO₄²⁻ → 3 ions

Step 1: Calculate ideal osmotic pressure if fully dissociated

$П_\text{ideal} = i_\text{ideal} \, C \, R \, T$

• $i_\text{ideal} = 3$ (fully dissociated)

$П_\text{ideal} = 3 \times 0.1 \times 0.082 \times 300$

$П_\text{ideal} = 7.38 \, \text{atm}$

Step 2: Determine van’t Hoff factor

$i_\text{observed} = \frac{П_\text{observed}}{C R T} = \frac{4.92}{0.1 \times 0.082 \times 300} = \frac{4.92}{2.46} \approx 2$

Step 3: Calculate percentage ionization

$i = 1 + (n - 1)\alpha$

• $n=3$ (number of ions)

$2 = 1 + (3 - 1)\alpha$

$2 = 1 + 2\alpha ⇒\alpha = \frac{1}{2} = 0.5$

$\text{Percentage ionization} = \alpha \times 100 = 50\%$