Question

If \( {A}_2{B} \) is 30% ionised in an aqueous solution, then the value of van't Hoff factor \( i \) is:

Hint:

The van't Hoff factor \( i \) is used to account for the effect of solute dissociation on colligative properties such as boiling point elevation and freezing point depression.

Answer 1

The van't Hoff factor \( i \) is the number of particles the compound dissociates into. For a compound that ionises, \( i \) is given by: \[ i = 1 + \alpha (n-1) \] where \( \alpha \) is the degree of ionisation and \( n \) is the number of ions produced per formula unit. Here, \( \alpha = 0.30 \) (since 30\% ionised) and for \( {A}_2{B} \), the ionisation produces 3 ions per formula unit. Thus, the van't Hoff factor is: \[ i = 1 + 0.30 \times (3-1) = 1 + 0.30 \times 2 = 1 + 0.60 = 1.60 \] So the value of \( i \) is \( 1.60 \times 10^{-1} \).