A) \[\frac{1}{5}th\] of the molar mass of \[KMn{{O}_{4}}\]
B) \[\frac{1}{6}th\] of the molar mass of \[KMn{{O}_{4}}\]
C) \[\frac{1}{3}rd\]of the molar mass of \[KMn{{O}_{4}}\]
D) \[\frac{1}{10}th\] of the molar mass of \[KMn{{O}_{4}}\]
Correct Answer: C
Solution :
In alkaline medium , \[2K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}+2KOH+{{H}_{2}}O\xrightarrow{{}}2\overset{+4}{\mathop{Mn{{O}_{2}}}}\,\]\[+4KOH+3[O]\] Decrease in oxidation number of \[Mn=(+7)-(+4)=3\] So, eq. wt. of \[KMn{{O}_{4}}=\frac{Mol.\,\,wt.}{3}\] Note However, in strongly alkaline medium (rarely used) following conversion occurs. \[2\,K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}+2KOH\xrightarrow{{}}2{{K}_{2}}Mn{{O}_{4}}+\]\[{{H}_{2}}O+[O]\] Change in oxidation number = 1 So, eq. wt. of \[KMn{{O}_{4}}=\frac{Mol.\,\,wt.}{1}\]
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