Question

100 mL of ${O_2}$ gas diffuses in 10 sec. 100mL of gas X diffuses in t sec. Gas X and time t can be:
a.) $S{O_2}$, 16 sec
b.) ${H_2}$, 2.5 sec
c.) CO, 10 sec
d.) He, 4 sec

Answer 1

Hint:The rate of diffusion of a gas is inversely proportional to the square root of molecular mass of the gas. This law is known as Graham’s law.

Complete step by step answer:
Diffusion is the gradual mixing of gases due to the motion of the component particles of the gas even in the absence of mechanical agitation such as stirring. The resultant mixture is a uniform mixture.
Graham law of diffusion states that the rate of diffusion or effusion of two gases is equal to the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. The mathematical representation of the law is:
$\dfrac{{{r_1}}}{{{r_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}}$
As in the question, same amount of gas is diffused, therefore we can directly consider rate of diffusion to be inversely proportional to the time taken
$\therefore \dfrac{{{t_1}}}{{{t_2}}} = \sqrt {\dfrac{{{M_1}}}{{{M_2}}}}$

Now, checking options,
(B) ${H_2}$, 2.5 sec
$\dfrac{{{t_1}}}{{{t_2}}} = \sqrt {\dfrac{{{M_1}}}{{{M_2}}}}$
$\dfrac{{10}}{{2.5}} = \sqrt {\dfrac{{32}}{2}}$
4 = 4

Hence, (B) is the correct option.
Other options can be checked similarly and the LHS and RHS won’t come the same.

Therefore, the correct answer is (B) ${H_2}$, 2.5 sec

Note: A student must note that if the volume of the gases diffused isn't the same, then the rate of diffusion isn’t directly equal to the inverse of time taken. In that case, the rate of diffusion is equal to the amount of gas diffused divided by the time taken.