KCET 2014 Chemistry Solved Paper

Given,
V (oxalic acid) =25cm3
W(NaOH)=0.064g
M (oxalic acid )=?
Since, moles of oxalic acid = mole of NaOH
V (oxalic acid) ×N (oxalic acid) =‌

w

M

(NaOH)
⇒N( oxalic acid )=‌

0.064×1000

25×40

=0.064
Normality = Molarity × Basicity
⇒ Molarity =‌

0.064

2

=0.32
Note: Basicity of oxalic acid is 2 .