Match the following:

List - I
List - II
A) Equivalent weight
of $K M n O_{4}$ in acidic medium
1) M/6
B) $K M n O_{4}$ in neutral medium
2) M/5
C) $K M n O_{4}$ in basic medium
3) M/3
D) Phosphorous acid
4) M/2

5) M/1

A-2, B-3, C-1, D-5

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A-2, B-3, C-5, D-4

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A-3, B-2, C-4, D-5

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A-3, B-2, C-5, D-4

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Solution

The correct option is A A-2, B-3, C-5, D-4
A. In acidic medium,
$M n^{+ 7} \to M n^{2 +}$
so, n-factor =5
so, equivalent weight = $\frac{M}{5}$

B. In neutral medium
$M n^{+ 7} \to M n^{+ 4}$
so, n-factor = 3
so, equivalent weight = $\frac{M}{3}$

C. In basic medium,
$M n^{+ 7} \to M n^{6 +}$
so, n-factor =1
so, equivalent weight = $\frac{M}{1}$

D. Phosphorous acid, have basicity =2,
so, equivalent weight = $\frac{M}{2}$

Option B is correct.

Suggest Corrections

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List - I	List - II
A) Equivalent weight of $K M n O_{4}$ in acidic medium	1) M/6
B) $K M n O_{4}$ in neutral medium	2) M/5
C) $K M n O_{4}$ in basic medium	3) M/3
D) Phosphorous acid	4) M/2
	5) M/1

Match the following:List - I List - IIA) Equivalent weight of KMnO4 in acidic medium1) M/6B) KMnO4 in neutral medium2) M/5C) KMnO4 in basic medium3) M/3D) Phosphorous acid4) M/25) M/1

Match the following:

List - I
List - II
A) Equivalent weight
of $K M n O_{4}$ in acidic medium
1) M/6
B) $K M n O_{4}$ in neutral medium
2) M/5
C) $K M n O_{4}$ in basic medium
3) M/3
D) Phosphorous acid
4) M/2

5) M/1