Question

0.004 M K₂SO₄ solution is isotonic with 0.01 M glucose solution. Percentage dissociation of K₂SO₄ is______ (Nearest integer)

Hint:

For isotonic solutions, use the equation \( i = 1 + (n-1) \) to find the number of particles and dissociation percentage.

Answer 1

Correct Answer: 75

For isotonic solution: \[ i(\text{glucose}) = i(\text{K}_2\text{SO}_4) \] \[ 0.01 = i(\text{K}_2\text{SO}_4) \times 0.004 \] \[ i(\text{K}_2\text{SO}_4) = \frac{0.01}{0.004} = 2.5 \] Now, for \( K_2SO_4 \): \[ i = 1 + (n-1) \] \[ 2.5 = 1 + (n-1) \] \[ n = 3 \text{ for } K_2SO_4 \] Percentage dissociation: \[ \alpha = \frac{3}{2} = 75\% \] Thus, the percentage dissociation of K_2SO_4 is 75%.