Question

Which of the following set of compounds does NOT demonstrate the law of multiple proportion?

• A. $H_2O,\ CO_2,\ CH_4$
• B. $H_2O,\ H_2O_2$
• C. $SO_2,\ SO_3$
• D. $NO,\ NO_2$

Hint:

The law of multiple proportion is applicable only when the same two elements form more than one compound.

Answer 1

The Correct Option is A

Step 1: State the law of multiple proportion.
The law of multiple proportion states that when two elements form more than one compound, the masses of one element which combine with a fixed mass of the other element are in the ratio of small whole numbers.
Step 2: Analyze each option.
(A) $H_2O,\ CO_2,\ CH_4$: These compounds do not consist of the same two elements. Since the law applies only when the same two elements form different compounds, this set does not demonstrate the law.
(B) $H_2O,\ H_2O_2$: Both compounds contain hydrogen and oxygen, and their mass ratios follow simple whole numbers. Hence, the law is obeyed.
(C) $SO_2,\ SO_3$: Sulfur and oxygen combine in simple whole-number ratios, satisfying the law.
(D) $NO,\ NO_2$: Nitrogen and oxygen also obey the law of multiple proportion.
Step 3: Conclusion.
Since option (A) does not contain compounds formed by the same two elements, it does not demonstrate the law of multiple proportion.