Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Solutions

Question:

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:

Options:

A-IV, B-III, C-I, D-II

A-IV, B-I, C-III, D-II

A-IV, B-III, C-II, D-I

A-I, B-II, C-III, D-IV

Correct Answer:

A-IV, B-III, C-I, D-II

Explanation:

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.