Question

1. What is the normality of KOH solution if 45.18 mL are required to neutralize 0.300 g of pure oxalic acid (H2C2O4·2H2O)?

Answer 1

$Moles =\frac{0.300} {126.07}=0.0024moles$

$45mL= 0.045L$

Molarity $= \frac{0.0024}{0.045}=0.053M$