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Osmotic Pressure

10.875 g of a...

Question

$10.875 g$ of a mixture of $N a C l$ and $N a_{2} C O_{3}$ was dissolved in water and the volume made up of $250 m L, 20 m L$ of this solution required $75.5 m L$ of $\frac{N}{10} H_{2} S O_{4}$ . Find out the percentage composition of the mixture.

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Solution

Only $N a_{2} C O_{3}$ will react with $H_{2} S O_{4}$
Applying $N_{1} V 1 (N a_{2} C O_{3}) \equiv N_{2} V_{2} (H_{2} S O_{4})$
$N_{1} \times 20 = 75.5 \times \frac{1}{10}$
$N_{1} = \frac{75.5}{20 \times 10} = 0.3775$
$N a_{2} C O_{3} 1 m o l . m a s s + H_{2} S O_{4} 2 g e q . \to N a_{2} S O_{4} + H_{2} O + C O_{2}$
Eq. mass of $N a_{2} C O_{3} = \frac{106}{2} = 53$
Mass of $N a_{2} C O_{3}$ present in $250 m L 0.3775 N$ solution'
$= \frac{N \times E \times V}{1000} = \frac{0.3775 \times 53 \times 250}{1000}$
$= 5.0018 g$
Mass of $N a C l = (10.875 - 5.0018) = 5.8732 g$
$N a_{2} C O_{3} = \frac{5.0018}{10.875} \times 100 = 45.99$ %
$N a C l = \frac{5.8732}{10.875} \times 100 = 54.0$ %.

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