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  4. 25 cm3 of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

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

KCETKCET 2014Some Basic Concepts of Chemistry

Solution:

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

$=2 \times 0.064$

$=0.032$