Calculate the amount of $K O H$ required to prepare 100 mL of 0.1 M solution.
(Atomic weight of $K = 10, O = 16, H = 1$ )

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Solution

Molarity= $\frac{Weight of KOH}{GMW of KOH} \times \frac{1000}{V o l (m l)}$
$[G M W$ of $K O H =$ Atomic weight of $K +$ atomic weight of $O +$ atomic weight of $H = 10 + 16 + 1 = 27]$
$\Rightarrow 0.1 = \frac{Weight of KOH}{27} \times \frac{1000}{100}$
Weight of $K O H = \frac{2.7}{10}$
Weight of $K O H = 0.27 g$

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