Question

1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper obtained was 1.098 g. In another experiment, 1.179 g of copper was dissolved in nitric acid and the resulting solution was evaporated to dryness. The residue of copper nitrate when strongly heated was converted into 1.4476 g of cupric oxide. The result of this process show:
(A) law of reciprocal proportion
(B) law of multiple proportion
(C) law of constant proportion
(D) none of these

Answer 1

The given problem shows two cases to be analysed; which would be easy to follow up and study all the given laws in the option.
- Knowing about the stated laws and solving the given illustration we can easily determine the required answer.

Complete Solution :
Let us study the given laws one by one:
Law of reciprocal proportion-
The law of reciprocal proportion states that if two different elements combine separately with a fixed mass of any third element; the ratio in which they do so is the same to the ratio of their masses, when those two elements combine with each other.
This law is also known as the law of equivalent proportions or law of permanent ratios.

Law of multiple proportion-
The law of multiple proportion states that if two elements react to form one or more compounds; then the ratios of the masses of the element with variable mass combining with the element of fixed mass can always be expressed as the ratio of small whole numbers.

Law of constant proportion-
The law of constant proportion states that the compound forming from the combination of some elements will form in the fixed ratio of those elements irrespective of their source or method of preparation.
This law is also known as the law of definite proportion or law of constant proportion.

Now, let us see the given illustration:
Here, we have two cases i.e.
- Case I-
1.375 g of cupric oxide contains 1.098 g copper. Thus, oxygen content will be 0.277 g.
Ratio of copper to oxygen = $\dfrac{1.098}{0.277}=3.9:1\simeq 4:1$
- Case II-
1.4476 g of cupric oxide contains 1.179 g copper. Thus, oxygen content will be 0.268 g.
Ratio of copper to oxygen = $\dfrac{1.179}{0.268}=4.3:1\simeq 4:1$
Therefore, we can say that this is the case of law of constant proportions.
So, the correct answer is “Option C”.

Note: Do note that we have other techniques too for analysing the given illustration by finding the percentage of oxygen in cupric oxide for both the cases. The equality in these percentages too can prove the same conclusion i.e. law of constant proportion.