7.95 gm oxide of copper on reaction with hydrogen shows a mass loss and the residue obtained weighs 6.35 gm. In another experiment, 19.05 gm of copper was dissolved in nitric acid and the copper nitrate produced was converted to oxide weighing 23.85 gm. This data illustrates:

Law of multiple proportion

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Law of mass conservation

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Law of reciprocal proportion

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Law of constant proportion

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Solution

The correct option is D Law of constant proportion
The ratio of Masses of oxide of $C u$ to $C u$ ,
In experiment $A = \frac{7.95}{6.35} = 1.25$
In experiment $B = \frac{23.85}{19.05} = 1.22$
It shows the law of constant proportion.

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Q. In an experiment, 4.90 g of copper oxide was obtained from 3.92 g of copper. In another experiment, 4.55 g of copper oxide gave, on reduction, 3.64 g of copper. Show with the help of calculations that these figures verify the law of constant proportions.

In a experiment, 4.90g of copper oxide was obtained from 3.92g of copper. In another experiment, 4.55g copper oxide gave on reduction 3.64g of copper. Show that these figures verify the law of constant proportion.

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7.95 gm oxide of copper on reaction with hydrogen shows a mass loss and the residue obtained weighs 6.35 gm. In another experiment, 19.05 gm of copper was dissolved in nitric acid and the copper nitrate produced was converted to oxide weighing 23.85 gm. This data illustrates:

The correct option is D Law of constant proportionThe ratio of Masses of oxide of Cu to Cu,In experiment A=7.956.35=1.25In experiment B=23.8519.05=1.22It shows the law of constant proportion.

The correct option is D Law of constant proportion
The ratio of Masses of oxide of $C u$ to $C u$ ,
In experiment $A = \frac{7.95}{6.35} = 1.25$
In experiment $B = \frac{23.85}{19.05} = 1.22$
It shows the law of constant proportion.