$3 .92 g .$ of Mohr salt $[mol .wt = 392]$ is present in $100 ml$ of an aqueous solution. The sulphate ion concentration of the resulting solution is

1 M

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2 M

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0.2 M

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0.02 M

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Solution

The correct option is C $0.2 M$
The chemical formula of Mohr salt is $(N H_{4})_{2} F e (S O_{4})_{2} .6 H_{2} O$
We are given:
Given mass of Mohr salt = $3.92 g$
Molar mass of Mohr salt = $392 g / m o l$
Volume of solution = $100 m L$
Concentration of mohr salt is $\frac{mole of mohr salt}{volume(L)}$
$= \frac{3.92 / 392}{100 / 1000} M = 0.1 M$
Now, the chemical equation for the ionization of mohr salt follows:
$(N H_{4})_{2} F e (S O_{4})_{2} .6 H_{2} O \to 2 N H_{4}^{+} + F e^{2 +} + 2 S O_{4}^{2 -} + 6 H_{2} O$
From the equation,
we can say that the concentration of sulfate ions $= (2 \times 0.1) = 0.2 M$

Option C is the correct answer.

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3.92g. of Mohr salt [mol.wt = 392] is present in 100ml of an aqueous solution. The sulphate ion concentration of the resulting solution is

$3 .92 g .$ of Mohr salt $[mol .wt = 392]$ is present in $100 ml$ of an aqueous solution. The sulphate ion concentration of the resulting solution is