Question

sulphur and oxygen are known to form oxide.the sulphur content in one of these is 50 percent while other is 40 percent . show that it illutrates the law of multiple proporation?
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Answer 1

Explanation:

In the first compound % of S = 51% , hence % of O will be = 49%

therefore if % of S = 1% , the % of O will be = 49 /51= 0.96.

In the second compound

the % of S = 41% , hence % of O will be = 59%

if % of S = 1% , % of O = 59/ 41 = 1.439

therefore the simple ratio between O: O in these oxide = 1.439/0.96 : 0.96/ 0.96 = 1.5 : 1 = 3:2 hence the law

Ratio of oxygen in oxides of sulphur is 3:2

Answer 2

20000is the answer of this question