Chemistry, asked by starshraddha3025, 1 year ago

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

Answers

Answered by tiwaavi
93
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. :-)



Answered by Kanak007vats
15

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