A solution containing $4.2$ g of KOH and $C a (O H)_{2}$ is neutralized by an acid. It consumes $0.1$ equivalent of acid, calculate the percentage composition of the sample.

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Solution

Let mass of $K O H$ be $a$ g and mass of $C a (O H)_{2}$ be $b$ g.
$∴ a + b = 4.2 ⟶ (1)$
For the given reaction:
eq. of $C a (O H)_{2} +$ eq. of $K O H =$ eq. of $H_{2} S O_{4}$
$\Rightarrow 2 \times$ moles of $C a (O H)_{2} + 1 \times$ mole of $K O H = 2 \times$ mole of $H_{2} S O_{4}$
$\Rightarrow 2 \times \frac{b}{74} + \frac{a}{56} = 2 \times \frac{4.9}{98}$
$\Rightarrow \frac{2 b}{74} + \frac{a}{56} = 0.1 ⟶ (2)$
Solving (1) and (2) we get $a = 1.47 g$ and $b = 2.73 g$ .
% of $K O H = \frac{1.47}{4.2} \times 100 = 35$ % and % of $C a (O H)_{2} = 100 - 35 = 65$ %

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