Question

Calculate the normality of H2C2O4.2H2O solution containing 1.575 g per 0.25 dm3. (At.wt. H=1, C=12, O=16)

Answer 1

Answer:

0.2N

Explanation:

Step I. Calculation of molarity of the sodium

Mass of oxalic acid =3.15g

Molar mass of oxalic acid =24+64+2+2×18=126u=126g mol−1

No. of moles of oxalic acid =(3.15g)(126g mol−1)

Volume of solution =250mL=2501000=0.25L

Molarity of solution (M)=No. of moles of oxalic acidVolume of solution in litres

=(3.15g)(126g mol−1)×(0.25L)

=0.1mol L−1=0.1M

Step II. Calculation of normality of the acid solution

Basicity of oxalic acid from the formula = 2

Equivalent mass of oxalic acid =Molecular massBasicity=(126g mol−1)2=(63 g equiv−1)

No. of equivalent of oxalic acid =Mass of oxalic acidEquivalent mass of oxalic acid=(3.15g)(63 g equiv−1)

=0.2equiv L−1=0.2N

Note : The normality of the acid solution can also be calculated with the help of relation :

Normality of acid = Molarity of acid × Basicity =0.1×2=0.2N