If \(A_2B \;\text{is} \;30\%\) ionised in an aqueous solution, then the value of van’t Hoff factor \( i \) is:
If \(A_2B \;\text{is} \;30\%\) ionised in an aqueous solution, then the value of van’t Hoff factor \( i \) is:
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Correct Answer: 1.6
Approach Solution - 1
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:
\(i = \frac{\text{total moles after ionisation}}{\text{initial moles of solute}} = \frac{1.6}{1} = 1.6\)
Approach Solution -2
For the compound \( A_2B \), dissociation in aqueous solution can be represented as:
\[ A_2B \rightleftharpoons 2A^+ + B^{2-} \]
If the degree of ionization (α) = 30% = 0.3, then out of 1 mole of \( A_2B \), 0.3 moles dissociate.
Step 2: Calculate total number of moles after dissociation.
Initial moles of \( A_2B \) = 1
Moles dissociated = 0.3
So, moles of undissociated \( A_2B \) = \( 1 - 0.3 = 0.7 \)
From 0.3 moles of dissociated \( A_2B \):
- Moles of \( A^+ \) formed = \( 2 \times 0.3 = 0.6 \)
- Moles of \( B^{2-} \) formed = 0.3
Total moles after dissociation = \( 0.7 + 0.6 + 0.3 = 1.6 \)
Step 3: Apply the van’t Hoff factor formula.
\[ i = \frac{\text{Total number of moles after dissociation}}{\text{Initial moles}} \] \[ i = 1.6 \]
Step 4: Final Result.
The van’t Hoff factor for 30% ionization of \( A_2B \) is:
\[ \boxed{i = 1.6} \]
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