Match the column-I (reaction) with column-II (percent purity)

	Column-I		Column-II
A)	1.72 g impure FeSO4 consumed 20 mL of 0.1M acidic KMnO4	P)	75% pure sample
B)	8.4 gm impure H2C2O4.2H2O consumed 0.1mole NaOH	Q)	79.67% pure sample
C)	9.84g FeSO4.(NH4)2SO4 .6H2O impure sample reduced 0.02 equivalent K2Cr2O7 acidic solution	R)	79.1% pure sample
D)	16.8 volume of 625ml of H2O2 reduced 75 gm impure KMnO4 in  acidic medium.	S)	88.37% pure sample

	A	B	C	D
1)	S	P	Q	R
2)	P	Q	R	S
S	P	R	Q	S
4)	S	Q	R	P

A) No. of moles of KMnO₄ = 20 × 0.1 × 5 = 10 
10 Meq KMnO₄ = 10 Meq FeSO₄
For FeSO₄ ; 10 = wt of FeSO₄/152 x 1000
wt of FeSO₄ = 1.52 g 
% purity of FeSO₄ = 1.52/1.72 x 100 = 88.37
B) 0.1 Mole NaOH = 
0.1 equivalent NaOH 
0.1 equivalents NaOH = 
0.1 equivalents H₂C₂O₄ .2H₂O
0.1 equivalents 
H₂C₂O₄ .2H₂O = 0.1 x 63 = 6.3g 
% purity of H₂C₂O₄ .2H₂O = 6.3/8.4 x 100 = 75
C) 0.02 equivalent K₂Cr₂O₇ = 0.02 
equivalents Mohr salt 
0.02 equivalents Mohr salt 
= 0.02 × 392 = 7.84 g
% purity of Mohr salt = 7.84/9.84 x 100 = 79.67
D) 1.87 eQ H₂O₂ = 1.87 eq KMnO₄ 
1.87 eQ KMnO₄ = 1.87 × 31.6 = 59.32 g 
% purity of KMnO₄ = 59.32/75 x 100= 79%