Question

If \(A_2B \;\text{is} \;30\%\) ionised in an aqueous solution, then the value of van’t Hoff factor \( i \) is:

Hint:

The van’t Hoff factor \( i \) depends on the degree of ionisation and the number of ions produced in the solution.

Answer 1

Correct Answer: 1.6

To determine the van’t Hoff factor (\(i\)) for a compound \(A_2B\) with 30% ionization, we must analyze the effect of ionization on \(i\).

Assume an initial quantity of 1 mole of \(A_2B\). During ionization, it dissociates according to:

\(A_2B \rightarrow 2A^+ + B^-\)

Complete ionization of 1 mole of \(A_2B\) would yield 2 moles of \(A^+\) and 1 mole of \(B^-\), totaling 3 moles of ions.

Given that ionization is only 30%:

• Ionized moles of \(A_2B\) = 0.3 moles
• Undissociated moles of \(A_2B\) = 0.7 moles

From the 0.3 moles of \(A_2B\) that ionize, the following ions are produced:

• Moles of \(A^+\) = 2 × 0.3 = 0.6 moles
• Moles of \(B^-\) = 0.3 moles

The total moles present after ionization are the sum of undissociated solute and the ions formed: 0.7 (undissociated) + 0.6 (\(A^+\)) + 0.3 (\(B^-\)) = 1.6 moles.

The van’t Hoff factor (\(i\)) is calculated as:

\(i = \frac{\text{total moles after ionisation}}{\text{initial moles of solute}} = \frac{1.6}{1} = 1.6\)