A) 392
B) 31.6
C) 278
D) 156
Correct Answer: A
Solution :
Mohr's salt is \[{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.FeS{{O}_{4}}.6{{H}_{2}}O.\] The equation is : \[5F{{e}^{2+}}MnO_{4}^{-}+8{{H}^{+}}\xrightarrow[{}]{{}}\]\[5F{{e}^{3+}}+M{{n}^{2+}}+4{{H}_{2}}O\] Total change in oxidation number of iron \[=(+3)-(+2)\] \[=+1\] So, equivalent wt. of Mohr's salt \[\text{=}\frac{\text{Mol}\text{. wt}\text{. of Mohr }\!\!'\!\!\text{ s salt}}{\text{1}}\] \[=\frac{392}{1}\] = 392You need to login to perform this action.
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