Question

14 g of N₂ and 36 g of ozone are at the same pressure and
temperature . Their volumes will be related as
(a) 2VN₂ = 3VO₃ (b) 3VN₂ = 2VO₃
(c) 3VN₂ = 4VO₃ (d) 4VN₂ = 3VO₃

Answer 1

The volumes will be related as (a) 2VN₂ = 3VO₃

1. PV=nRT => V =nRT/P => V =kn where k is a constant as R T and P are constants in this case.

2. Let X be the moles of N2 and Y be that of O3

3. So, 14/28 =1/2 moles of N2 and 36/48= 3/4 moles of O3

4. Thus X/2 = 3Y/4

=> 2X = 3Y

=> 2VN₂ = 3VO₃

Answer 2

The volumes will be related as 3V(N) = 4 VO3

Option (C) is correct.

Explanation:

As we know that:

• Mass of nitrogen N = 14 g
• Mass of ozone = 36 g

V(N) = nRT / P = 14 / 14 x RT / P

V(N) = RT / P

V O3 = 36 / 48 x RT / P

V(N) / V (O3) = RT / P / 3/4 RT /V

3V(N) = 4 VO3

Thus the volumes will be related as 3V(N) = 4 VO3

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