A) 78.4
B) 70
C) 37
D) 40
Correct Answer: A
Solution :
[a] m-eq.\[FeS{{O}_{4}}{{(N{{H}_{4}})}_{4}}S{{O}_{4}}.6{{H}_{2}}O\] \[=m-eq.\text{ }of\,KMn{{O}_{4}}\] (n=l) \[\frac{W}{392}\times 1\times 1000=0.1\times 50\] Hence, % purity of Mohr's salt \[=\frac{1.96}{2.5}\times 100=78.4%\] \[W=1.96g\]You need to login to perform this action.
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