Question

how to prepare 100 ml 0.1 N ackalic acid​

Answer 1

Answer:

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Explanation:

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Answer 2

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Molarity = weight*1000/molecular weight*vol

Oxalic acid = COOH.COOH.2H2O

Molecular weight = 126g/mol

n factor=2

weight = ?

Molarity M = 0.1

volume = 100mL

Therefore 0.1 = x * 1000/ 126*100

x = 1.26

In normality

Normality = weight*1000/Equivalent weight*vol

Equivalent weight = 63g/mol

0.1 = x *1000/63*100

x = 0.63g