$x$ g of Fe $_{2}$ (SO $_{4}$ ) $_{3}$ was dissolved in water to prepare $1$ L of its aqueous solution. Upon analysis, it was found that each ml of the above solution contains $1.2 \times 10^{- 4}$ moles of $S O {_{4}}^{2 -}$ ions. Find the value of $x$ .

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Solution

The correct option is A 16
1 mole of ferric sulphate dissociates in water to give 3 moles of sulphate ions.

Hence, $\frac{x}{400}$ moles of ferric sulphate present in 1 L will dissociate to form $\frac{3 x}{400}$ moles of of sulphate ions.

But moles of sulphate ions present in 1 ml are $1.2 \times 10^{- 4}$ .

Hence, $\frac{3 x}{400 \times 1000} = 1.2 \times 10^{- 4}$

$x = 16 g$ .

16 g of ferric sulphate was dissolved in 1 L of solution.

Hence, option A is correct.

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