Question

1.325 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 250 ml.on titration,25 no of this solution neutralise 20 no of a solution of sulphuric acid.how much water should be added to 450 of this acid solution to make it exactly n/12?

Answer 1

Answer:

Explanation:

=> Here, It is given that,

Mass of anhydrous sodium carbonate = 1.325 g

Volume of solution = 250 ml

Equivalent mass = 52.9

=> Normality of Na₂CO₃ = $\frac{Mass of Na_2CO_3} {Equivalent mass}* \frac{1000}{Volume of solution}$                                

$=\frac{1.325}{52.9} *\frac{ 1000}{250}$

=0.1 N

=>According to the law of equivalence,

N₁V₁ (Na₂CO₃) = N₂V₂(H₂SO₄)

N₂ = N₁V₁/V₂

=0.1×25/20

=0.125 N

Here, 450 ml of 0.125 N H₂SO₄ is diluted to make it exactly N/12.

Thus, proportion of water to be added is:

450 * 0.125 = 1/12 * V

V = 450 * 0.125 * 12

V=675 ml

Thus, proportion of water to be added:

= 675−450= 225 ml

=> Hence, 225 ml of water should be added to 450 of this acid solution to make it exactly n/12

Answer 2

Answer:

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