Answer

A-S; B-P; C-Q; D-R

Explanation

A) No. of moles of `KMnO_(4)=20xx0.1xx5=10`
10 Meq `KMnO_(4)=10` Meq `FeSO_(4)`
For `FeSO_(4),10=("wt of "FeSO_(4))/(152)xx1000`
wt of `FeSO_(4)=1.52g`
% purity of `FeSO_(4)=(1.52)/(1.72)xx100=88.37`
B) 0.1 Mole `NaOH=0.1` equivalent NaOH
0.1 equivalents NaOH
= 0.1 equivalents `H_(2)C_(2)O_(4).2H_(2)O`
0.1 equivalents
`H_(2)C_(2)O_(4).2H_(2)O=0.1xx63=6.3g`
% purity of `H_(2)C_(2)O_(4).2H_(2)O=(6.3)/(8.4)xx100=75`
C) 0.02 equivalent `K_(2)Cr_(2)O_(7)=0.02`
equivalents Mohr salt
0.02 equivalents Mohr salt
`=0.02xx392=7.84g`
% purity of Mohr salt = `(7.84)/(9.84)xx100=79.67`
D) 1.87 eQ `H_(2)O_(2)=1.87` ecq `KMnO_(4)`
1.87 eQ `KMnO_(4)=1.87xx31.6=59.32g`
% purity of `KMnO_(4)=(59.32)/(75)xx100=79%`