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Equivalent Mass

1.20 g sample...

Question

$1.20 g$ sample of $N a_{2} C O_{3}$ and $K_{2} C O_{3}$ was dissolved in water to form $100 m L$ of a solution. $20 m L$ of this solution required $40 m L$ of $0.1 N H C l$ for complete neutralization. Calculate the weight of precipitate in the mixture if another $20 m L$ of this solution is treated with excess of $B a C l_{2}$ .

A

0.394 g

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B

0.596 g

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C

0.604 g

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D

0.115 g

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Solution

The correct option is A $0.394 g$
Solution of $N a_{2} C O_{3} a n d K_{2} C O_{3}$ gives ppt. of $B a C O_{3} w i t h B a C l_{2}$ .
$M e q . o f N a_{2} C O_{3} + M e q . o f K_{2} C O_{3} i n 20 m L = M e q . o f B a C O_{3}$
$\Rightarrow M e q . o f H C l f o r 20 m L m i x t u r e = M e q . o f B a C O_{3}$
$\Rightarrow M e q . o f B a C O_{3} = 40 \times 0.1 = 4$
$\frac{W_{B a C O_{3}}}{M_{B a C O_{3}}} \times 2 \times 1000 = 4$
$\frac{W_{B a C O_{3}}}{197} \times 2 \times 1000 = 4$
$∴ W_{B a C O_{3}} = 0.394 g$

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Equivalent Mass

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