Practicing Success
Match List I with List II
Choose the correct answer from the options given below: |
A-III, B-I, C-II, D-IV A-I, B-III, C-IV, D-II A-III, B-II, C-I, D-IV A-III, B-I, C-IV, D-II |
A-III, B-I, C-IV, D-II |
The correct answer is option 4. A-III, B-I, C-IV, D-II. 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 \(= 1 + 2 × 0.40 = 1.8\) |