`25cm^(3)` of oxalic acid completely neutralised 0.064g of soldium hydroxied. molarity of the oxalic acid solution is

Options

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

Answer

D

Explanation

`underset(COOH)overset(COOH)| + 2NaOH to underset(COONa)overset(COONa)| + 2H_(2)O`
Moles of 0.064 g of NaOH `=0.064/40 = 0.0016`
Since 2 moles of NaOH react with 1 mol of oxalic acid,
Moles of oxalic acid `=0.0016/2 = 8 xx 10^(-4)`
Volume of solution = 25 mL
Molarity of oxalic acid solution `=(8 xx 10^(-4))/25 xx 1000 = 0.032` M.