Question

1.7 g of ammonium salt was heated with excess of NAOH the ammonia released in the process neutralises 100 cc solution of N/5 H2SO4 what is precentage of ammonia in the salt,
(1) 20%
(2) 35%
(3) 1.7%
(4) 40%​
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Answer 1

The weight of ammonium salt = 1.7 g

100 cc or 100 ml solution of N/5 H₂SO₄ was neutralised in the process of ammonia release.

To find: % of ammonia in the salt

In 1 Litre, the normality will be,

= [Weight / equivalent weight] * Volume

= [1.7 / 17.03] * 1 …… [since equivalent weight of ammonia is 17 u]

= 0.1 N

Let the % of ammonia in the salt be “x”. Also, let the normality and the volume of ammonia be denoted as “N₁” & “V₁” and of H₂SO₄ be denoted as “N₂” & “V₂” respectively.

By rearranging the normality equation we will calculate the value of “x” i.e.,  

x = [N₂V₂ / N₁V₁] * 100  

or, x = [{1/5 * 100} / {0.1 * 1000}] * 100 = 20%

Thus, the percentage of ammonia in the salt is 20%.

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