Match List I with List II
|
LIST I Salts |
LIST II van't Hoff Factor |
| A. Mohr's salt | I. 3 |
| B. Blue Vitriol | II. 8 |
| C. Lime slake | III. 2 |
| D. Alum | IV. 5 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is option 1. A-IV, B-III, C-I, D-II.
|
LIST I Salts |
LIST II van't Hoff Factor |
| A. Mohr's salt | IV. 5 |
| B. Blue Vitriol | III. 2 |
| C. Lime slake | I. 3 |
| D. Alum | II. 8 |
Let us go through each salt and its corresponding van't Hoff factor:
A. Mohr's Salt: The formula of Mohr's salt is \((NH_4)_2Fe(SO_4)_2.6H_2O\). Mohr's salt dissociates to five ions:
\((NH_4)_2Fe(SO_4)_2.6H_2O \longrightarrow 2NH_4^+ + Fe^{2+} + 2SO_4^{2-}\)
Since, the number of ions formed is five. Thus the van't Hoff factor, \(i = 5\).
B. Blue vitriol: The formula of blue vitriol is \(CuSO_4 . 5H_2O\). Blue vitriol dissociates to two ions:
\(CuSO_4.5H2O \longrightarrow Cu^{2+} + SO_4^{2-}\)
Since, the number of ions formed is two. Thus the van't Hoff factor, \(i = 2\).
C. Lime slake: The formula of lime slake is \(Cu(OH)_2\). Lime slake dissociates into three ions:
\(Cu(OH)_2 \longrightarrow Cu^{2+} + 2OH^-\)
Since, the number of ions formed is three. Thus the van't Hoff factor, \(i = 3\).
D. Alum:he formula of alum is \(K_2SO_4Al_2(SO_4)_3.24H_2O\). alum dissociates into eight ions:
\(K_2SO_4Al_2(SO_4)_3.24H_2O \longrightarrow 2K^+ + 2Al^{3+} + 4SO_4^{2-}\)
Since, the number of ions formed is eight. Thus the van't Hoff factor, \(i = 8\).
Thus the correct match is option 1. A-IV, B-III, C-I, D-II.