Question

25 cm3 of oxalic acid completely neutralised
0.064 g of sodium hydroxide. Molarity of the
oxalic acid solution is
fa) 0.064
(b) 0.045
(c) 0.015
(d) 0.032​

Answer 1

$\large\underline{\underline{\sf Given:}}$

• Volume of Oxalic Acid (V) = 25cm³

• Mass of NaOH (m) = 0.064

$\large\underline{\underline{\sf To\:Find:}}$

• Molarity of Oxalic Acid (M) .

$\large\underline{\underline{\sf Solution:}}$

We know ,

${\boxed{\sf \frac{Volume×Normlity}{1000}=\frac{Mass\:of\:NaOH}{Molecular\:mass\:of\:NaOH} }}$

$\large\implies{\sf \dfrac{25×N}{1000}=\dfrac{0.064}{40} }$

$\large\implies{\sf N=0.064 }$

Then,

$\large{\boxed{\sf Molarity (M)=\frac{Normality}{n-factor} }}$

$\large\implies{\sf M=\dfrac{0.064}{2} }$

$\large\implies{\sf M=0.032 }$

$\Large\underline{\underline{\sf Answer:}}$

•°• Molarity of the oxalic acid solution is 0.032

Answer 2

Molarity of oxalic acid solution is 0.032.

We know that 1cm^3=1 ml

So 25 cm^3=25ml

Volume of oxalic acid =25ml

Weight of NaOH=40grams

Given weight=0.064

Moles =given weight/molecular wt

Moles=0.064/40=16×10^-4

Molarity of NaOh=moles /volume(l)

M=moles×1000/volume(ml)

M=16×10^-4×1000/25

Molarity of NaOh=0.064 mole/ml

Now when oxalic acid completely neutralize NaOH so their molarity will be equal but for NaOH molarity will be equal to normality .

But for oxalic acid

Molarity=Normality/nfactor

Nfactor for oxalic acid =2

Molarity of oxalic acid =0.064/2

Molarity of oxalic acid=0.032M