Question

$25{cm}^3$ of oxalic acid completely neutralised $0.064g$ of sodium hydroxide. Molarity of the oxalic acid solution is :

• A. $0.045$ M
• B. $0.032$ M
• C. $0.064$ M
• D. $0.015$ M

Answer 1

The correct option is B $0.032$ M

$(COOH{)}_2+2NaOH\to (COONa{)}_2+2H_{2}O$

Moles of $NaOH=\frac{0.064}{40}=1.6\times {10}^{-3}moles$

Moles of $(COOH{)}_2=\frac{1}{2}molesofNaOH=0.8\times {10}^{-3}moles$

$M=\frac{moles}{V}=\frac{0.8\times {10}^{-3}}{25\times {10}^{-3}}=0.032M$