Copper oxide was prepared by two different methods. In case, 1.75 g of the metal gave 2. 19 g of oxide. In the second case, 1.14 g of the metal gave 1.43 g of the oxide, show that the given data illustrate the law of constant proportions.

Copper oxide was prepared by two different methods. In case, 1.75 g of

1.	Copper oxide was prepared by two different methods. In case, 1.75 g of the metal gave 2. 19 g of oxide. In the second case, 1.14 g of the metal gave 1.43 g of the oxide, show that the given data illustrate the law of constant proportions.
Answer» In case I, mass of copper = 1.75 g Mass of copper oxide = 2.19 g % of copper in the oxide `= ("Mass of copper")/("Mass of copper oxide")xx 100` `=(1.75)/(2.19)xx100 = 79.9 %` `therefore %` of oxygen `= 100 - 79.9 = 20.1 %` In case II, mass of copper = 1.14 g Mass of copper oxide = 1.43 g % of copper in the oxide `=(1.14)/(1.43)xx100 = 79.7 %` % of oxygen = 100 - 79.7 = 20.3 %. Thus, copper oxide prepared by any of the given methods contain copper and oxygen in the same proportion by mass (with the experimental error). Hence, it proves the law of constant proportions.

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Copper oxide was prepared by two different methods. In case, 1.75 g of the metal gave 2. 19 g of oxide. In the second case, 1.14 g of the metal gave 1.43 g of the oxide, show that the given data illustrate the law of constant proportions.

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