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

Question

400 mg of capsule contains 100 mg of ferrous fumerate. The percentage of iron present in the capsule is approximately (A) 8.2 (B) 25% (C) 16% (D) unpredictable

Tardigrade Admin · Accepted Answer

Mol. mass of Fe(CHCOO)2=170 Fe in 100 mg of Fe(CHCOO)2=(56 × 100/170) =32.9 mg Total Fe in 400 mg of capsule =32.9 mg ∴ percentage of Fe in capsule =(32.9 × 100/400) =8.2

Q. $400 m g$ of capsule contains $100 m g$ of ferrous fumerate. The percentage of iron present in the capsule is approximately

8.2

25%

16%

unpredictable

Mol. mass of $F e (C H COO)_{2} = 170$
Fe in $100 m g$ of $F e (C H COO)_{2} = \frac{56 \times 100}{170}$
$= 32.9 m g$
Total Fe in $400 m g$ of capsule $= 32.9 m g$
$∴$ percentage of Fe in capsule
$= \frac{32.9 \times 100}{400}$
$= 8.2$

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

8.2

25%

16%

unpredictable

Mol. mass of Fe(CHCOO)2​=170 Fe in 100mg of Fe(CHCOO)2​=17056×100​ =32.9mg Total Fe in 400mg of capsule =32.9mg ∴ percentage of Fe in capsule =40032.9×100​ =8.2

Q. $400 m g$ of capsule contains $100 m g$ of ferrous fumerate. The percentage of iron present in the capsule is approximately

Mol. mass of $F e (C H COO)_{2} = 170$
Fe in $100 m g$ of $F e (C H COO)_{2} = \frac{56 \times 100}{170}$
$= 32.9 m g$
Total Fe in $400 m g$ of capsule $= 32.9 m g$
$∴$ percentage of Fe in capsule
$= \frac{32.9 \times 100}{400}$
$= 8.2$