Question

Explain law of reciprocal of water of proportion using the following data
1) ammonia contains 82.35% N
2) water contains 88.9%N
3) Nitrogen sesquioxide contains 63.15%O

Answer 1

Answer:

Given,

NH_3NH

3

contains 82.35% of N_2N

2

and 17.65% of H_2H

2

H_2OH

2

O contains 88.90% of O_2O

2

and 11.10% of H_2H

2

NO_3NO

3

contains 63.15% of O_2O

2

and 36.85% of N_2N

2

In ammonia, 17.65g of H combine with N = 82.35g

Therefore, 1g of H combine with N = 82.35/17.65 = 4.67g

In water, 11.10g of H combine with O = 88.90

1g of H combine with O = 88.90/11.10=8.01g

Ratio of the masses of N and O which combine well with fixed mass (is equal to 1g) of H =4.67: 8.01 = 1:1.72

In N_2ON

2

O ratio of masses of N and O which combine with each other = 36.85: 63.15 = 1:1.72.

Thus, two ratios are the same. Hence, it proves the law of reciprocal proportions