Question

$25 \, cm^3$ of oxalic acid completely neutralised $0.064\, g$ of sodium hydroxide. Molarity of the oxalic acid solution is

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

Answer 1

Answer: (B)

0.032

Answer 2

Moles of oxalic acid = Moles of $NaOH$
$V \times N =\frac{W}{M}$
$\frac{25}{1000} \times N =\frac{0.064}{40}$
$N =\frac{0.064 \times 1000}{25 \times 40}=0.064$
$\because$ Molarity of oxalic acid $=$ Basicity $\times$ Normality

$=2 \times 0.064$
$=0.032$