Question

What weight of Na2CO3 of 95% purity would be required to
neutralize 45.6 mL of 0.235 N acid?
a 60 gm
b. 80 gm
C. 0.5978 g
d. 0.9 gm
​

Answer 1

Given:

Purity of Na2CO3 = 95%

Volume of acid = 45.6ml

Normality = 0.235 N

To find :

Weight of Na2CO3.

Solution:

• For complete neutralization,

        Meq. of Na2CO3 = Meq. of H2SO4

        W / ( 106/2 ) x 1000= 45.6 x 0.235

                                    W = 0.5679g

• As per the question, 95g of pure Na2CO3 is to be taken, then

        Weighed sample = 100g

• Now, for 0.5679g of Na2CO3,

        Weighed sample = (100x0.5679)/95

                                     = 0.5978 g

• This weight of Na2CO3 will be 0.5978 g.