Question

Volume of $0.1 \,M\, NaOH$ needed for the neutralisation of $20\, mL$ of $0.05\, M$ oxalic acid is

• A. $10\,mL$
• B. $15\,mL$
• C. $20\,mL$
• D. $30\,mL$

Answer 1

Answer: (C)

$20\,mL$

Answer 2

Me of $NaOH$ = Me of oxalic acid ; $0.1 \times V = 20 \times 0.05 \times 2$ $\Rightarrow V=20\, mL$