A) 85.36%
B) 14.64%
C) 58.63%
D) 26.14%
Correct Answer: B
Solution :
[b] Let the amount of the \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] in the mixture be x g, then amount of \[KMn{{O}_{4}}\] will be (0.5-x)g \[\therefore \left( \frac{x}{49}+\frac{0.5-x}{31.6} \right)=\frac{100\times 0.15}{1000}\] Where 49 is Eq. wt. of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] and 31.6 is Eq. wt. of \[KMn{{O}_{4}}.\] On solving, we get x = 0.073 g Percentage of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] \[=\frac{0.0732\times 100}{0.5}=14.64\]%You need to login to perform this action.
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