Question

Volume of 0.1 M NaOH needed for the neutralisation
of 20 ml of 0.05 M oxalic acid is :​

Answer 1

$\huge{\mathcal{BONJOUR \: !!}}$

Answer:

N1V1 = N2V2

V1 x 0.1 = 0.05 x 20

V1 = 1/0.1

V1 = 10 ml

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