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:
Given data:
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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