JEE Advanced 2021 Paper 2 · Q12 · Mole Concept & Stoichiometry
The value of $y$ is _____.
Reveal answer + step-by-step solution
Correct answer:18.75
Solution
Mass of Fe in 250 mL solution = moles $\times$ molar mass = $1.875 \times 10^{-2} \times 56 = 1.05$ g.
Percentage by weight: $y = \frac{1.05}{5.6} \times 100 = 18.75 \%$.
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