Question

An aqueous solution of
6.3g
oxalic acid dihydrate is made up to 250ml.The volume of 0.1N NaOH
required to completely neutralize
10ml of this solution is??

Answer 1

_______________________


Equivalent mass of:-


$C_{2}H_{2}O_{4}.2H_{2}O$ = $\frac{126}{2} = 63$


Normality of oxalic acid = $\frac{6.3}{63} \times \frac{1000}{250} = 0.4N$


$\bold{We\:know\:that,}$


$N_{1}V_{1}$ = $N_{2}V_{2}$


$\bold{Where,}$


$N_{1}$ is normality of Oxalic acid.


$V_{1}$ is volume of Oxalic acid.


$N_{2}$ is normality of Sodium hydroxide.


$V_{2}$ is volume of Sodium hydroxide.


$\bold{Therefore,}$


$0.4 \times 10 = 0.1 \times V_{2}$


$V_{2}$ = ${40\:ml}$


$\bold{Therefore,}$


${40\:ml}$ of NaOH is required to completely neutralize ${10\:ml}$ of oxalic acid.


_______________________

Answer 2

This question involves concept of volumetry

We can get the answer using the formula

N1V1=N2V2 ie. Principle of neutralization

Here

Oxalic acid is a dibasic acid as it can release two H+

Hence Molarity = (6.3 x 1000)/( 126 x 250) = 0.2

=> normality = 0.2 x 2 = 0.4

For req volume of NaOH

0.4 x 10 = 0.1 x V2

=> V2 = 40 ml

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