Question

A sample of na2co3.h2o weighning 0.62 g is added to 100 ml of 0.1n h2so4. the resulting solution will be

Answer 1

Hello Dear.

According to the question,Here is the Reaction Involved in this Process.

Na₂CO₃ + H₂SO₄ --------→ NA₂SO₄ + CO₂ + 2H₂O

Now, we have to find the Normality of Na₂CO₃ .

Equivalent weight of Na₂CO₃.H₂O is  62 g.

Given weight of Sodium carbonate (Na₂CO₃) = 0.62 g.

Now by Formula,

W =  $\frac{N . E . V}{1000}$

0.62 = ( N x 62 x 100 ) / 1000

N = 0.62 x 1000 / 62 x 100
N = 620/6200
N= 0.1 

Hence the Normality of Na₂CO₃ and H₂SO₄ is Same , 0.1 N .
So the resulting solution will be Neutral.

Hope it Helps. :-)

Answer 2

please see the attachment

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