QJEEM21Ch06ed: JEE Main 2021 Chemistry - Class 11 JEEM21Ch06ed Solution: Consider titration of NaOH solution versus 1.25 M oxalic acid…

Average burette reading = Volume of NaOH solution (V₁)

$= {{4.5 + 4.5 + 4.4 + 4.4 + 4.4} \over 5}$

= 4.44 mL

Strength of NaOH solution = S₁(M) (say) = S₁(N)

Volume of oxalic acid solution (V₂) = 10 mL

Strength of oxalic acid solution (S₂) = 1.25 M = 1.25 $\times$ 2 N

So, ${V_1}{S_1} = {V_2}{S_2}$ ( $\because$ Law of equivalence)

$\Rightarrow {S_1} = {{{V_2}{S_2}} \over {{V_1}}} = {{10 \times (1.25 \times 2)} \over {4.44}} = 5.63$ N

$\simeq 6M = 6M$