A) 158
B) 52.67
C) 31.6
D) 49
Correct Answer: C
Solution :
Key Idea: Equivalent weight of \[\text{KMn}{{\text{O}}_{\text{4}}}\] \[\text{=}\frac{\text{formula}\,\text{weight}\,\text{of}\,\text{KMn}{{\text{O}}_{\text{4}}}}{\text{change}\,\text{in}\,\text{oxidation}\,\text{number}}\] \[\text{2KMn}{{\text{O}}_{4}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}\] \[+2MnS{{O}_{4}}+3{{H}_{2}}O+5[O]\] or \[\underset{+7}{\mathop{Mn{{O}_{4}}^{-}}}\,+16{{H}^{+}}\xrightarrow{{}}\underset{+2}{\mathop{M{{n}^{2+}}}}\,+8{{H}_{2}}O\] \[\therefore \] Equivalent weight of \[\text{KMn}{{\text{O}}_{\text{4}}}\]in acidic \[medium=\frac{158}{5}=31.6\]
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