An aqueous solution is prepared by dissolving pure crystals of Mohr’s salt $F e S O_{4} {(N H_{4})}_{2} S O_{4} .6 H_{2} O$ in water. Density of the above solution is $1.2 g / m L$ and the solution contains $11.833 % F e S O_{4} {(N H_{4})}_{2} S O_{4}$ by weight. Then find the mass of the salt dissolved in $400 m L$ volume of solution.
(Molar masses: $F e = 56, S = 32, N = 14, H = 1, O = 16$ )

Question

An aqueous solution is prepared by dissolving pure crystals of Mohr’s salt $F e S O_{4} {(N H_{4})}_{2} S O_{4} .6 H_{2} O$ in water. Density of the above solution is $1.2 g / m L$ and the solution contains $11.833 % F e S O_{4} {(N H_{4})}_{2} S O_{4}$ by weight. Then find the mass of the salt dissolved in $400 m L$ volume of solution.
(Molar masses: $F e = 56, S = 32, N = 14, H = 1, O = 16$ )

Accepted Answer

Since molarity is defined as moles of solute per litre of solution, it is always recommended to consider $1.0 L$ of solution to determine molarity when the density is given.
Therefore, mass of one litre of the above solution
$= v o l u m e \times d e n s i t y = 1000 m L \times 1.2 g / m L = 1200 g$
Mass of $F e S O_{4} {(N H_{4})}_{2} S O_{4}$ present in $1.0 L$ of above solution
$= Mass of 1.0 L solutio n \times \frac{30}{100} = 1200 \times \frac{11.833}{100} = 142 g$
Mass of $F e S O_{4} {(N H_{4})}_{2} S O_{4}$ present in $1.0 L$ solution
$= \frac{142}{284} = 0.5$
$\Rightarrow M o l a r i t y (M) = 0.5 M$
Moles of anhydrous salt in $400 m L$ solution
$= 0.5 \times 0.40$
$= 0.2$
Moles of hydrated salt dissolved = Moles of the anhydrous salt in solution
Moles of hydrated salt dissolved for $400 m L$ of solution = 0.2
Mass of hydrated salt dissolved $= 0.2 \times 392 = 78.4 g$

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