Match List I with List II List I List II A. K_2SO_4 (aq) with 60% dis…

The correct answer is option 4. A-III, B-I, C-IV, D-II.

List I	List II
A. \(K_2SO_4\) (aq) with \(60\%\) dissociation	III. \(i = 2.2\)
B.  \(K_3[Fe(CN)_6]\) (aq) with \(90\%\) dissociation	I. \(i = 3.7\)
C. \(AlCl_3\) (aq) with \(80\%\) dissociation	IV. \(i = 3.4\)
D. \(K_2HgI_4\) (aq) with \(40\%\) dissociation	II. \(i = 1.8\)

A. \(K_2SO_4\) (aq) with \(60\%\) dissociation

\(K_2SO_4\) is an \(A_2B\) type of electrolyte, for which van't Hoff factor is given by \(1 + 2\alpha \)

Where,

\(\alpha = \frac{\% \, \ dissociation}{100}\)

Since the % dissociation is 60%

Thus,

\(\alpha = \frac{60}{100} = 0.60\)

So, van't Hoff factor \(= 1 + 2 × 0.60 = 2.2\)

B.  \(K_3[Fe(CN)_6]\) (aq) with \(90\%\) dissociation:

\(K_3[Fe(CN)_6]\) is an \(A_3B\) type of electrolyte, for which van't Hoff factor is given by \(1 + 3\alpha \)

Where,

\(\alpha = \frac{\% \, \ dissociation}{100}\)

Since the % dissociation is 90%

Thus,

\(\alpha = \frac{90}{100} = 0.90\)

So, van't Hoff factor \(= 1 + 3 × 0.90 = 3.7\)

C. \(AlCl_3\) (aq) with \(80\%\) dissociation:

\(AlCl_3\) is an \(A_3B\) type of electrolyte, for which van't Hoff factor is given by \(1 + 3\alpha \)

Where,

\(\alpha = \frac{\% \, \ dissociation}{100}\)

Since the % dissociation is 80%

Thus,

\(\alpha = \frac{80}{100} = 0.80\)

So, van't Hoff factor \(= 1 + 3 × 0.80 = 3.4\)

D. \(K_2HgI_4\) (aq) with \(40\%\) dissociation:

\(K_2HgI_4\) is an \(A_2B\) type of electrolyte, for which van't Hoff factor is given by \(1 + 2\alpha \)

Where,

\(\alpha = \frac{\% \, \ dissociation}{100}\)

Since the % dissociation is 40%

Thus,

\(\alpha = \frac{40}{100} = 0.40\)

So, van't Hoff factor \(= 1 + 2 × 0.40 = 1.8\)