Calculate the percentage composition of iron in ferric oxide, $F e_{2} O_{3}$ (atomic mass of $F e = 56)$

70%

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56%

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87%

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35%

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Solution

The correct option is A 70%
i) Molecular weight of $F e_{2} O_{3}$ $= (56 \times 2) + (16 \times 3) = 160 g$
Molar mass of Fe in 1 mole of $F e_{2} O_{3}$ $= 56 \times 2 = 112 g$
Therefore %age composition of Fe in $F e_{2} O_{3}$ $= \frac{m o l a r m a s s o f F e}{M o l e c u l a r w e i g h t o f F e_{2} O_{3}} \times 100 = \frac{112}{160} \times 100 = 70$ %
ii) Molecular weight of Urea( $H_{2} N C O N H_{2}$ ) $= 2 + 14 + 12 + 16 + 2 + 14 = 60 g$
Molar mass of N in 1 mole of Urea $= 2 \times 14 = 28 g$
Therefore %age composition of N in Urea $= \frac{m o l a r m a s s o f N}{M o l e c u l a r w e i g h t o f H_{2} N C O N H_{2}} \times 100 = \frac{28}{60} \times 100 = 46.6$ %

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