A) 33%
B) 25%
C) 14%
D) 8%
Correct Answer: D
Solution :
Molecular weight of \[{{(CHCOO)}_{2}}Fe=170\] Fe present in 100mg of \[{{(CHCOO)}_{2}}Fe\] \[=\frac{56}{170}\times 100mg=32.9mg\] This is present in 400mg of capsule % of Fe in capsule \[=\frac{32.9}{400}\times 100=8.2\].You need to login to perform this action.
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