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 find the van’t Hoff factor (\(i\)) for the compound \(A_2B\) which is 30% ionised, we need to understand how ionisation impacts \(i\).

Assume initially we have 1 mole of \(A_2B\). Upon ionisation, it dissociates as follows:

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

This means, for complete ionisation, from 1 mole of \(A_2B\), we would get 2 moles of \(A^+\) and 1 mole of \(B^-\), equating to 3 moles of ions in total.

However, it is only 30% ionised. Therefore:

• Moles of \(A_2B\) ionised = 0.3 moles
• Moles of \(A_2B\) undissociated = 0.7 moles

From 0.3 moles of \(A_2B\) ionised, ions produced are:

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

Total moles after ionisation = moles undissociated + moles of ions = 0.7 + 0.6 + 0.3 = 1.6 moles

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

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