How many grams of sodium thiosulphate ( $N a_{2} S_{2} O_{3} .5 H_{2} O$ ) is required to prepare $250 m l$ of $0.1 M$ ?

10.2 g

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9.3 g

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8.5 g

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6.2 g

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Solution

The correct option is C 6.2 g
Molarity= $\frac{w e i g h t}{M o l a r m a s s} \times \frac{1000}{V (m l)}$

$\Rightarrow 0.1 M = \frac{W e i g h t}{248} \times \frac{1000}{250}$

$\Rightarrow W e i g h t = \frac{248 \times 0.1}{4} = 6.2 g$

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