A) 3.47 g
B) 12.38 g
C) 1.23 g
D) 34.76 g
Correct Answer: A
Solution :
[a] Normality of ferrous amm sulphate \[=\frac{3.92\times 1000}{392\times 100}=0.1\](Eq. wt of FAS is 392) \[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\] \[20\times 0.1=18\times {{N}_{2}}\] \[{{N}_{2}}=0.111\] 1 g ev. of \[KMn{{O}_{4}}=31.6g\] 0.111 g ev. of \[KMn{{O}_{4}}=31.6\times 0.111=3.5g.\]You need to login to perform this action.
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