Question #0d32e

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Question #0d32e

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Answer 1 · 2016-10-18T08:49:57

In first oxide $Oxygen \to 20 % and Metal \to 80 %$ $"Oxygen"->20% and "Metal"->80%$

Here Ratio of

$Oxygen : Metal = 1 : 4$ $"Oxygen":"Metal"=1:4$

In 2nd oxide $Oxygen \to 11.1 % and Metal \to 88.9 %$ $"Oxygen"->11.1% and "Metal"->88.9%$

Here Ratio of

$Oxygen : Metal = 1 : 8$ $"Oxygen":"Metal"=1:8$

So ratio of masses of metal combining with same mass of Oxygen in two oxides will be $= 4 : 8 = 1 : 2$ $=4:8=1:2$ which is a ratio of simple whole number .So the law of multiple proportion is followed by the given data

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