`17.4%` (wt.`//`vol.) `K_(2)(SO_(4))` solution at `27^(@)C` isotonic to `5.85%` (wt.`//`vol.) `NaCl` solution at `27^(@)C`. IF `NaCl` is `100%` ionised, what is `%` ionisation of `K_(2)SO_(4)` in aq. solution?

Explanation

To solve the problem, we need to find the percentage ionization of \( K_2SO_4 \) in an aqueous solution, given that it is isotonic with a \( 5.85\% \) (wt/vol) \( NaCl \) solution.

### Step-by-Step Solution:

1. **Understand the Given Data**:
- We have a \( 17.4\% \) (wt/vol) \( K_2SO_4 \) solution.
- We have a \( 5.85\% \) (wt/vol) \( NaCl \) solution.
- Both solutions are isotonic at \( 27^\circ C \).

2. **Calculate the Mass of Solutes**:
- For \( K_2SO_4 \):
\[ \text{Mass of } K_2SO_4 = 17.4 \text{ g in } 100 \text{ mL} \]
- For \( NaCl \):
\[ \text{Mass of } NaCl = 5.85 \text{ g in } 100 \text{ mL} \]

3. **Calculate Molarity**:
- **Molar mass of \( K_2SO_4 \)**:
\[ K: 39.1 \times 2 + S: 32.1 + O: 16 \times 4 = 174.3 \text{ g/mol} \]
- **Molarity of \( K_2SO_4 \)**:
\[ \text{Molarity} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]
\[ = \frac{17.4}{174.3 \times 0.1} = 1 \text{ M} \]

- **Molar mass of \( NaCl \)**:
\[ Na: 23 + Cl: 35.5 = 58.5 \text{ g/mol} \]
- **Molarity of \( NaCl \)**:
\[ = \frac{5.85}{58.5 \times 0.1} = 1 \text{ M} \]

4. **Calculate Van't Hoff Factor (i)**:
- For \( K_2SO_4 \):
- It dissociates into \( 2K^+ + SO_4^{2-} \), so \( n = 3 \).
- The Van't Hoff factor \( i \) is given by:
\[ i = 1 + (n - 1) \cdot \alpha = 1 + 2\alpha \]

- For \( NaCl \):
- It dissociates into \( Na^+ + Cl^- \), so \( n = 2 \).
- Since \( NaCl \) is 100% ionized, \( \alpha = 1 \):
\[ i = 1 + (2 - 1) \cdot 1 = 2 \]

5. **Set Up the Isotonic Condition**:
- Since the solutions are isotonic, their osmotic pressures are equal:
\[ \pi_{K_2SO_4} = \pi_{NaCl} \]
\[ i_{K_2SO_4} \cdot C_{K_2SO_4} = i_{NaCl} \cdot C_{NaCl} \]
\[ (1 + 2\alpha) \cdot 1 = 2 \cdot 1 \]
\[ 1 + 2\alpha = 2 \]

6. **Solve for \( \alpha \)**:
\[ 2\alpha = 1 \implies \alpha = 0.5 \]

7. **Calculate Percentage Ionization**:
- Percentage ionization of \( K_2SO_4 \):
\[ \text{Percentage Ionization} = \alpha \times 100 = 0.5 \times 100 = 50\% \]

### Final Answer:
The percentage ionization of \( K_2SO_4 \) in aqueous solution is **50%**.