A) 33%
B) 25%
C) 14%
D) 8%
Correct Answer: D
Solution :
Molecular mass of \[{{(CHCOO)}_{2}}Fe=170\] \[\therefore \]In 100 g \[{{(CHCOO)}_{2}}Fe,\]iron present \[=\frac{56}{170}\times 100\,mg=32.9\,mg\] Since, this quantity of Fe is present in 400 mg of capsule, \[\therefore \] % of Fe in capsule\[=\frac{32.9}{400}\times 100=8.2\,%\]You need to login to perform this action.
You will be redirected in
3 sec