`25 cm^(3)` of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

Options

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

Answer

D

Explanation

`underset("1 mol")((COOH)_(2))+underset(2 " moles")(2NaOH) rarr(COONa)_(2)+2H_(2)O`
Mol. mass of NaOH `=40 "g mol"^(-1)`
No. of g moles in 0.064 g of NaOH
`=(0.064)/(40)=0.0016`
No. of moles of oxalic acid `= (0.0016)/(2)=8xx10^(-4)`
Volume of solution (in L) `=(25)/(1000)`
Hence, molarity `= ("No. of moles of solute")/("Volume of solution(in L)")`
`=8xx10^(-4)xx(1000)/(25)=0.032M`