In an experiment 1.288g of copper oxide was obtained from 1.03g of copper. In another experiment 3.672g of copper oxide obtain from the 2.938g of copper. Show that these figures verify the law of constant proportion.

Open in App

Solution

Dear Student,
First experiment

Copper oxide = 1.288 g

Copper left = 1.03 g

Oxygen present = 1.288 - 1.03 = 0.258 g

Percentage of oxygen in CuO = (0.258 × 100) / 1.288 = 19.98 ~ 20 %

Second experiment

Copper oxide = 3.672 g

Copper left = 2.938 g

Oxygen present = 3.672 - 2.938 = 0.734 g

Percentage of oxygen in CuO = (0.734 × 100) / 3.672 = 20.03 ~ 20 %

So, it is verified that these figures verify the law of constant proportion.
Best wishes!

Suggest Corrections

Similar questions

Q. In an experiment, 1.288 g of copper oxide was obtained from 1.03 g of copper. In another experiment, 3.672 g of copper oxide gave, on reduction, 2.938 g of copper. Show 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.

Q. (a) Define the Law of constant proportion by giving
one example.

(b) In an experiment, 1.288 g of copper oxide was obtained from 1.03 g of Cu. In another experiment, 3.672 g of copper oxide gave a reduction of 2.938 g of copper. Which law of chemical combination can be derived by this example?

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.

Q. In an experiment

1.288 g

of copper oxide was obtained from

1.03 g

of copper and

0.258 g

of oxygen. Calculate the ratio of copper and oxygen in the sample?

Join BYJU'S Learning Program