Question

400 mg of a capsule contains 100 mg of ferrous fumarate. The percentage of iron present in the capsule is approximately.

• A. 8.2%
• B. 25%
• C. 16%
• D. unpredictable

Answer 1

Answer: (A)

Molecular formula of ferrous fumarate is $H{e}^{+}=\frac{1}{\lambda }=R_H.Z^2\left[\frac{1}{n_2^2}-\frac{1}{n_2^2}\right]$ , molar mass is 170 $=R_H{(2)}^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$ 100 mg of $=4R_H\times \frac{3}{16}=\frac{3}{4}{R}_H$ contains $\text{s }>\text{ p}>\text{ d}>\text{ f}$ $CsCl$ Total Fe in 400 mg of capsule $C{l}^{-}$ $\text{C}{\text{s}}^{+}$ percentage of Fe in capsule $\frac{1}{8}\times 8=1$