A) 0.02
B) 0.05
C) 0.04
D) 0.07
Correct Answer: B
Solution :
Change in oxidation number\[=(+7)-(+2)=5\] \[\therefore \] Normality of solution\[=5\times 0.04=0.20\text{ }N\] Volume = 250 mL \[\therefore \] Number of equivalents of\[KMn{{O}_{4}}\] \[=0.20\times \frac{250}{1000}=0.05\]You need to login to perform this action.
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