If A₂B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is _______ × 10⁻¹.

Solution

Chemistry ma’am · 3 steps

Count the ions A₂B breaks into first, then i = 1 + α(n-1) does the rest.

1. Count the ions. A₂B → 2A⁺ + B²⁻, so one formula unit gives n = 3 ions.
2. Particles per unit. Start with 1 mol and let a fraction α ionise. You are left with 1 - α of undissociated A₂B plus nα of ions. The total is 1 + α(n-1).

   i = 1 + α(n-1) = 1 + 0.3 × (3-1) = 1.6

3. So. i = 1.6 = 16 × 10⁻¹, so the blank is 16.

The common slip: 13. Counting A₂B as giving only 2 ions makes n = 2, so i = 1.3 and the blank becomes 13.

The idea to keep. For partial dissociation, i = 1 + α(n-1), where n is the total number of ions from one formula unit. Count every ion, including repeated ones like the two A⁺.

Answer from the NTA final answer key. Solution by PrepWiser.