Question

Solve Q8 properly with explanation​

Answer 1

Given:

• Purity of Sodium carbonate, Na₂CO₃ = 95%

• Volume of acid = 45.6 ml

• Normality = 0.235 N

To find:

The Weight of Sodium carbonate, Na₂CO₃

Solution:

• Volume of acid  * Normality (H₂SO₄)= Number of Moles of Na₂CO₃

(H₂SO₄ is also called as Sulfuric acid or sulphuric acid)

or

• 45.6 ml X 0.235 N= Number of Moles of Na₂CO₃

Now,

• The equivalent Weight of Sodium carbonate, Na₂CO₃ is

=Molecular Weight /2

(Here, molecular weight is divided by 2 as it reacts with 2 moles of monoprotic acid)

=106/2

=0.5679 g

Now,  

• Weighed sample is 100 g for 95 g of pure Sodium carbonate, Na₂CO₃.  

Therefore,

• 0.5679 g of Sodium carbonate, Na₂CO₃ will be  

• Weighed sample = (100 X 0.5679) / 95

                                      = 0.5978 g

• Therefore the weight of Sodium carbonate, Na₂CO₃ will be 0.5978 g.