Question

A metal forms two oxides. The higher oxide contains 80% metal. 0.72g of the lower oxide gave 0.8g of higher oxide when oxidized. Then the ratio of the weight of oxygen that combines with the fixed weight of the metal in the two oxides will be:
(a) 2:3
(b) 2:1
(c) 4:5
(d) 3:2

Answer 1

Hint: Use law of multiple proportions in this question, i.e. the proportions of elements present in chemical compounds can be expressed in small whole number ratios.

Complete step by step answer:
Let us apply law of multiple proportions.
According to the question, when we oxidize 0.72g of lower oxide, it gives 0.8g of higher oxide.
Therefore, lower oxide + oxygen $\to$higher oxide
0.72 g of lower oxide on oxidation gives 0.8 g higher oxide.
So, 1 g of lower oxide on oxidation will give $\dfrac{0.8}{0.72}$g higher oxide.
Therefore, for 100 g higher oxide, $\dfrac{\text{0}\text{.72 x 100}}{\text{0}\text{.8}}$g of lower oxide is required = 90 g.
From this, we can say that 90g of lower oxide contains the same amount of metal as 100g higher oxide.
According to the question, higher oxide contains 80% metal.
So, 100g higher oxide will contain 80g metal.
Now we can conclude that – both oxides contain 80g metal.
Higher oxide combines with, 100g – 80g = 20g oxygen
Lower oxide combines with, 90g – 80g = 10g oxygen
So, the ratio is = 20:10 = 2:1
Therefore, the answer is – option (b).

Additional Information:
The law of multiple proportions was formulated by John Dalton.

Note: Law of multiple proportions states that “if two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers”.