Question

illustrate the law of Definite proportion 0.32 gram of sulphur on burning in air produced 224 ml of a SO2illustrate the law of Definite proportion 0.32 gram of sulphur on burning in air produced 224 ml of a SO2 a metal sulphide react with a mineral acid to give Sulphur Dioxide which contain 50% of Sulphur and 50% of oxygen solve this question ​

Answer 1

Answer:  In both the cases the ratio between S and O in SO2 remains constant.

Explanation:

There are two separate experiments given in the question.

1) The first experiment is where 0.32 grams of sulphur is burned in air to produce 224 ml of $SO_2$.

2) In the second experiment where 224ml of  $SO_2$ is produced when metal sulphide reacts with mineral acid to give $SO_2$ which contains 50% sulphur and 50% oxygen.

In first experiment:

224ml of $SO_2$ contains 0.32g of S

which means , 22400 ml contains 32 g of S

Also, we know that molecular mass of Oxygen is 16 u.

and  $SO_2$  has 2 molecules of oxygen , so  $SO_2$ has 32 g of Oxygen.

So, the ration between S and O in $SO_2$ is 32:32 = 1:1

In second experiment:

The  $SO_2$ contains 50 % of S so it contains 50% of O also

so the ratio between S:O  = 50:50 = 1:1

So, in both the cases the ratio between S and O in  $SO_2$ remains constant.

Answer 2

Thus in both the cases the ratio between S and O in SO2 remains constant.

i.e. 1 : 1

Explanation:

1st experiment:

• 224 ml of SO2 contains 0.32 g of  sulfur.
• Thus 22400 ml contains 32 g of sulfur.
• The ratio between Sulfur and Oxygen in SO2 is 32:32 = 1:1

2nd experiment:

• Similarly in second experiment ,the SO2 contains 50 % of S so it contains 50% of O also.
• Hence the ratio between S:O  = 50 : 50 = 1:1
• Thus in both the cases the ratio between S and O in SO2 remains constant.