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5g sample contain only $N a_{2} C O_{3} & N a_{2} S O_{4} .$ This sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralizes 20 mL of $0.1 M H_{2} S O_{4} .$ Calculate the % of $N a_{2} S O_{4}$ in the sample.

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57.6

36.2

42.4

5.76

answer is B.

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Detailed Solution

Only $N a_{2} C O_{3}$ reacts with $H_{2} S O_{4}$

$N a_{2} C O_{3} + H_{2} S O_{4} \overset{}{\to} N a_{2} S O_{4} + H_{2} C O_{3}$

$m -$ moles of $N a_{2} C O_{3} = m -$ moles of $H_{2} S O_{4} = 20 \times 0.1 = 2$

$m -$ moles of $N a_{2} C O_{3}$ in 250 ml solution. $= \frac{250}{25} \times 2 = 20$

Wt. of $N a_{2} C O_{3} = 20 \times 106 \times 10^{- 3} = 2.12 g$

$% o f N a_{2} C O_{3} = \frac{2.12}{5} \times 100 = 42.4$

$% o f N a_{2} C O_{3} = 100 - 42.4 = 57.6$

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