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 \[\text{= 5 }\!\!\times\!\!\text{ 0}\text{.04 = 0}\text{.20}\,\text{N}\] Volume = 250 mL \[\therefore \] Number of equivalents of \[\text{KMn}{{\text{O}}_{\text{4}}}\] \[=0.20\times \frac{250}{1000}\] \[=0.05\]You need to login to perform this action.
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