A METAL FORMS TWO OXIDES . THE HIGHER OXIDE CONTAINS 80% METAL.0.72gram OF THE LOWER OXIDE GAVE 0.8gram OF THE HIGHER OXIDE WHEN OXIDIZED. SHOW THAT THE DATA ILLUSTRATES THE LAW OF MULTIPLE PROPORTIONS?

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Solution

Given:
Amount of metal in higher oxide = 80%
0.72g of lower oxide give=0.8g of higher oxide
Now,
Metal in higher oxide = 80% of 0.8
= 0.64g
Oxygen in higher oxide = 20% of 0.8
= 0.16g
Now, amount of oxygen in lower oxide = (0.8-0.72)g
= 0.08g
Also, metal in lower oxide = (0.72-0.08)g
= 0.64g
Now, according to law of multiple proportion, ratio of amount of oxygen in higher oxide to the amount of oxygen in lower oxide should be a whole number, i.e.
= 0.16/0.08
= 2
Hence, this is in accordance with law of multiple proportion.

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Similar questions

A metal forms two oxides, the higher oxide contains 80% metal, 0.72g of the lower oxide gave 0.8g of higher oxide on heating. Show that the data illustrates the law of multiple proportions?

Q. A metal forms two oxides. The higher oxide contains

80 %

metal.

0.72 g

of the lower oxide gave

0.8 g

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:

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