Q. A 400 mg iron capsule contains 100 mg of ferrous fumarate, $(CHCOO)_{2}Fe.$ the percentage of iron present in it is approximately
A) 33%
B) 25%
C) 14%
D) 8%
Explanation
(d)
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 $= \frac{32.9}{400} \times 100 = 8.2\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.