Question

In an experiment, the weight of Sulphur in Sulphur dioxide was 50%. In another experiment, 3.2 grams of Sulphur on burning produced 6.4 grams of Sulphur dioxide. Show that the results prove the law of constant proportions.
[Hint: Prove that the ratio of weights of Sulphur and Oxygen is 1: 1 in both the experiments]
please answer it and answer correctly​

Answer 1

Answer:

In the first experiment:-

Weight of copper oxide, (CuO)=1.288g

Weight of copper =1.03g

Weight of oxygen = Wt. of copper oxide − Wt. of copper =1.288−1.03=0.258g

∴

Wt. of oxygen

Wt. of copper

=

0.258

1.03

≈

1

4

In the second expriment:-

Weight of copper oxide (CuO)=3.672g

Weight of copper =2.938g

Weight of oxygen = Wt. of copper oxide − Wt. of copper =3.672−2.938=0.734g

∴

Wt. of oxygen

Wt. of copper

=

0.734

2.938

≈

1

4

The proportion of elements of copper and oxygen in the reactions is same.

Hence law of constant proportion is proved.