Question

Consider titration of $NaOH$ solution versus 1.25 $M$ oxalic acid solution. At the end point following burette readings were obtained.

$\left(i\right)4.5\text{mL}\left(ii\right)4.5\text{mL}\left(iii\right)4.4\text{mL}\left(iv\right)4.4\text{mL}\left(v\right)4.4\text{mL}$

If the volume of oxalic acid taken was $10.0\text{mL}$. Then the molarity of the $NaOH$ solution is $M$. (Rounded-off to the nearest integer)

Answer 1

Since, $4.4\text{mL}$ is repeated thrice, it is taken as the final value.
Equivalent of $NaOH$ = Equivalent of Oxalic acid
$\left[NaOH\right]\times 1\times 4.4=5/4\times 2\times 10$
$\left[NaOH\right]=100/4\times 4.4=25/4.4=5.68$
$\text{Nearest integer = 6 M}$.