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Percentage Composition

4 g of a mix...

Question

$4 g$ of a mixture of $N a_{2} S O_{4}$ and anhydrous $N a_{2} C O_{3}$ were dissolved in pure water and volume made up to $250 m L$ . $20 m L$ of this solution required $25 m L$ of $N / 5 H_{2} S O_{4}$ for complete neutralisation. Calculate the percentage composition of the mixture.

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Solution

By applying normality formula,
$N_{1} V_{1} = N_{2} V_{2} N_{1} \times 20 = 0.2 \times 25 N_{1} = 0.25$
In terms of molarity $M_{1} = \frac{N_{1}}{2} = \frac{0.25}{2} = 0.125 M$
If molar of $N a_{2} S O_{4} = \frac{x}{142}, N a_{2} C O_{3} = \frac{4 - x}{106}$
Now, total moles,
$0.03125 = \frac{x}{142} + \frac{4 - x}{106} = \frac{106 x + 568 - 142 x}{142 \times 106}$
$470.375 = 106 x + 568 - 142 x$
$97.625 = 36 x ⟹ x = \frac{97.625}{36} = 2.7$
$∴ 2.7 g (M g S O_{4}), (4 - 2.7) = 1.3 g (M a_{2} C O_{3})$

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