Question

Which of the following has Van't Hoff factor (i) 1.82? (m = 0.01)

• A. \(\text{K}_2\text{SO}_4\)
• B. \(\text{MgSO}_4\)
• C. \(\text{NaCl}\)
• D. \(\text{HCl}\)
• E. \(\text{KCl}\)

Hint:

Higher charges on ions lead to greater deviation from the ideal Van't Hoff factor. A 2:2 salt like \(\text{MgSO}_4\) will have a much lower experimental \(i\) than a 1:1 salt like \(\text{NaCl}\) at the same concentration.

Answer 1

The Correct Option is C

Step 1: Concept of Van't Hoff Factor
Van't Hoff factor (\(i\)) represents the number of particles formed after dissociation of an electrolyte. Due to interionic attraction, the experimental value of \(i\) is always less than the ideal value.
Step 2: Ideal Van't Hoff Factor
Calculate ideal \(i\) (number of ions formed):
\[ \begin{aligned} \text{K}_2\text{SO}_4 &\rightarrow 2\text{K}^+ + \text{SO}_4^{2-} \quad (i = 3)
\text{MgSO}_4 &\rightarrow \text{Mg}^{2+} + \text{SO}_4^{2-} \quad (i = 2)
\text{NaCl} &\rightarrow \text{Na}^+ + \text{Cl}^- \quad (i = 2)
\text{HCl} &\rightarrow \text{H}^+ + \text{Cl}^- \quad (i = 2)
\text{KCl} &\rightarrow \text{K}^+ + \text{Cl}^- \quad (i = 2) \end{aligned} \] Step 3: Comparison with Given Value
Given \(i = 1.82\), which is close to 2 but slightly less.
This indicates a 1:1 electrolyte showing slight ion-pair formation.
- \(\text{K}_2\text{SO}_4\): Eliminated (\(i \approx 3\))
- \(\text{MgSO}_4\): Shows strong ion pairing due to \(2^+/2^-\) charges, so \(i\) is much lower (\(\approx 1.5\))
- \(\text{NaCl}, \text{KCl}, \text{HCl}\): All are 1:1 electrolytes
Among these, standard experimental data at low concentration (\(m = 0.01\)) shows:
\[ \text{NaCl} \approx 1.82} \] Step 4: Final Answer
\[ \boxed{\text{NaCl}} \]