Copper oxide was prepared by 2 different methods in 1 case 1-75 g of metal give 2-19 gram of oxide in case 21-14 gram of metal give 1-43 gram of oxide-how we can illustrate this according to the law of constant proportion

Solution : 1st method : weight of Cu : 1.75 g weight of CuO : 2.19 g Ratio of weight of Cu and CuO = 1.75 : 2.19 = 1 : 0.7990 ≈ 5 : 4 2nd method : weight ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Chemistry

Law of Constant Proportions

Copper oxide ...

Question

Copper oxide was prepared by 2 different methods. In case 1, 1.75 g of metal gives 2.19 grams of oxide. In case 2, 1.14 grams of metal gives 1.43 grams of oxide. How we can illustrate this according to the law of constant proportion

Open in App

Solution

In case 1, 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×100 = $\frac{1.75}{2.19} \times 100 = 79.9$ %
∴% of oxygen =100−79.9=20.1%
In case 2, the mass of copper = 1.14 g
Mass of copper oxide = 1.43 g
% of copper in the oxide = $\frac{1.14}{1.43} \times 100 = 79.7$ %
% of oxygen = 100 - 79.7 = 20.3 %.
Thus, copper oxide prepared by any of the given methods contains copper and oxygen in the same proportion by mass (with the experimental error). Hence, it proves the law of constant proportions.

Suggest Corrections

109

Similar questions

Q. Copper oxide was prepared by two different methods. In case one, 1.75 g of the metal gave 2.19 g of the oxide. In the second case, 1.14 g of the metal gave 1.43 g of the oxide. This is explained by which of the following laws?

Q. Two oxides of certian metal were separetly heated in a current of hydrogen untill constant weights were obtained . The.water produces in each case was carefully collected and weighed . 2 grams of each oxide gave respectively 0.2517 grams and 0.4526 grams of water .show that these results establosh the law of multiple proportions.

Q. Copper forms two oxides following law of variable proportions. One gram of each oxide in hydrogen gas gave

0.799 g

and

0.888 g

of the metal respectively. Give the composition of these oxides.

2.8 g

of calcium oxide prepared by heating limestone were found to contain

0.8 g

of oxygen. When one gram of oxygen was treated with calcium,

3.5 g

of calcium oxide was obtained. Show that the results illustrate the law of definite proportions.

Join BYJU'S Learning Program