Question

A metal combines with oxygen to form two oxides having the following composition
i) 0.398 gram of metal oxide I contains 0.318 gram of metal
ii) 0.716 gram of metal oxide II contains 0.636 gram of metal. So that the above data agrees with the law of multiple proportions.

Answer 1

Wt of oxygen in first oxide = 0.398 - 0.318 = 0.080 g Wt of oxygen in second oxide = 0.716 - 0.636 = 0.080 g   For fixed mass of oxygen, (0.080g), metal forms oxides in the ratio = 0.318:0.636 = 1:2 Since 1:2 is a simple ratio, it agrees with the law of multiple proportions.

Answer 2

Hello there,

● Law of multiple proportions -

• It states that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

● Illustration -

(i) In metal oxide I,

Mass of oxide = 0.398 g

Mass of metal = 0.318 g

Mass of oxygen = 0.08 g

Ratio of O/M = 0.08/0.318 = 0.252

(ii) In metal oxide II,

Mass of oxide = 0.716 g

Mass of metal = 0.636 g

Mass of oxygen = 0.08 g

Ratio of O/M = 0.08/0.636 = 0.126

Simple ratio of Oxygen in two metal compounds will be 0.252:0.126 = 2:1.

Thanks dear...