If a certain oxide of nitrogen weighing 0.11g gives 56 ml of nitrogen and another oxide of nitrogen weighing 0.15 g gives the same volume of nitrogen (both at STP), show that these results support the law of multiple proportions.

Answer 1

22400 ml of N2 = 28 gm so,

56 ml of N2 = 28 x 56/ 22400 = 0.07 gm

Here weight of oxide (A) =0.11 gm

After reduction the weight = 0.07gm

Loss of weight = 0.11-0.07 = 0.04gm (It is the amount of oxygen)--1.

Similarly weight of oxide(B) =0.15 gm

Weight after reduction = 0.07 gm

Hence, in case (A)

0.07 gm of nitrogen combines with 0.04 gm of oxygen

so, weight of nitrogen which would combine with 0.08 gm of oxygen should be equal to

= 0.07 x 0.08/ 0.04 = 0.14 gm--.---2

Here,in both the cases 1 and 2 the weight of nitrogen which combines with fixed amount of oxygen ( 0.04 gm) are 0.07 gm and 0.14 gm which are in the simple ratio of 2:1

Hence the oxides follow the law of multiple proportions.

Answer 2

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Answer 3

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Answer 4

Sir maine nitrogen ka mass fixed kr diya aur oxygen ka ratio nikala 1:2 aa rha hai kya mera answer shi hai kya please reply