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Q.
$400\, mg$ of capsule contains $100 \,mg$ of ferrous fumerate. The percentage of iron present in the capsule is approximately
Some Basic Concepts of Chemistry
Solution:
Mol. mass of $Fe(CHCOO)_{2}=170$
Fe in $100 \,mg$ of $Fe(CHCOO)_{2}=\frac{56 \times 100}{170}$
$=32.9\,mg$
Total Fe in $400\,mg$ of capsule $=32.9\,mg$
$\therefore $ percentage of Fe in capsule
$=\frac{32.9 \times 100}{400}$
$=8.2$