G. whose vapour Escample: If 360cm of H₂
diffuse in 1 hour, how long
will it take 600am of gas
density is
25 diffuse under the same
condition - If e relative mole
culer mass of H2 is a

8.33 hours

Explanation:

In order to solve this problem, we must apply Graham's law of diffusion in gases. Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its vapour density. For two gases we can write;

R1/R2=√d2/d1

Where;

R1= rate of diffusion of hydrogen

R2= rate diffusion of unknown gas

d1= vapour density of hydrogen

d2= vapour density of the unknown gas

Volume of hydrogen gas = 360cm^3

Time taken for hydrogen gas to diffuse= 1 hour =3600 secs

R1 = 360 cm^3/3600 secs = 0.1 cm^3 s-1

Vapour density of unknown gas = 25

Vapour density of hydrogen = 1

Substituting values,

0.1/R2 = √25/1

0.1/R2 = 5/1

5R2 = 0.1 × 1

R2 = 0.1/5

R2= 0.02 cm^3s-1

Volume of unknown gas = 600cm^3

Time taken for unknown gas to diffuse= volume of unknown gas/ rate of diffusion of unknown gas

Time taken for unknown gas to diffuse= 600/0.02

Time= 30,000 seconds or 8.33 hours

it would be a gas formation