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Question

Find the amount of 98% pure Na2CO3 required to prepare 5 litres of 2 N solution.

A
54.8 g impure Na2CO3
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B
540.8 g impure Na2CO3
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C
40.8 g impure Na2CO3
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D
340.8 g impure Na2CO3
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Solution

The correct option is B 540.8 g impure Na2CO3
no. of equivalents = normality × volume
=5×2 =10
Also, no. of equivalents =weightequivalentweight
Therefore, weight×2molecularweight=10....... [n factor for Na2CO3=2]
Weight =5x molecular weight = 5x 106 =530
Sample is 98% pure
amount required = 530×10098 =540.8g impure Na2CO3

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