Mohr's salt, FeSO4.(NH4)2SO4.6H2O, is used as a reductant (Fe^2+ -> Fe^3+). Using Fe=56, N=14, S=32, O=16, H=1, find the mass (in g, two decimals) required to prepare 500 mL of 0.0840 N solution.

JEE Main levelNumerical answerDifficulty 2 of 3eq180

Mohr's salt, $FeSO X_{4} \cdot (NH X_{4}) X_{2} SO X_{4} \cdot 6 H X_{2} O$ , is used as a reductant ( $Fe X^{2 +} \to Fe X^{3 +}$ ). Using $Fe = 56$ , $N = 14$ , $S = 32$ , $O = 16$ , $H = 1$ , find the mass (in $g$ , two decimals) required to prepare $500 mL$ of $0.0840 N$ solution.

Numerical answer type, answer in g, tolerance 0.03.

Molar mass

= 56 + 2 (18) + 2 (96) + 6 (18) = 392

; n-factor

= 1

, so equivalent mass

= 392

. Equivalents needed

= 0.0840 \times 0.500 = 0.0420

. Mass

= 0.0420 \times 392 = 16.464 g