Question

Hydrogen diffuses six times faster than gas A. The molar mass of gas A is:
A.$72{\text{ }}g/mole$
B.${\text{6 }}g/mole$
C.${\text{24 }}g/mole$
D.${\text{36 }}g/mole$

Answer 1

Rates of diffusion and effusion both are elaborated on the basis of Graham's law which states that under similar conditions of pressure and temperature the rates of diffusion (or effusion) of gases are inversely proportional to the square roots of their densities (or molecular masses). $\mathop 6\nolimits^2 = \dfrac{{\mathop M\nolimits_A }}{{\mathop 2\nolimits^2 }}$
So , $\mathop M\nolimits_A$will be equals to
$6 \times 6 \times 2 \times 2{\text{ }} = {\text{ }}72g/mole$

Complete step by step answer:
Let us see the relation between rate of diffusion and molar mass:-
$r\; \propto \dfrac{1}{{\sqrt M }}$ ( where r is rate of diffusion and M is molar mass) $\mathop r\nolimits_H$:- rate of diffusion of hydrogen
$\mathop r\nolimits_A$ :- rate of diffusion of gas A
Now rH = 6× rA ( as given in question)
MH ( molar mass of hydrogen , H2 = 2)
Now $\mathop M\nolimits_A$ = molar mass of gas A that we have to find.
Applying graham’s law of diffusion
$\dfrac{{\mathop r\nolimits_H }}{{\mathop r\nolimits_A }} = \dfrac{{\sqrt {\mathop M\nolimits_A } }}{{\sqrt {\mathop M\nolimits_H } }}$
Putting the values in this equation
$\dfrac{{\mathop {6r}\nolimits_H }}{{\mathop r\nolimits_A }} = \dfrac{{\sqrt {\mathop M\nolimits_A } }}{2}$
Squaring both sides it becomes

Our required option is A that is $72{\text{ }}g/mole$

Note:
Diffusion is the process of intermixing of two or more gases irrespective of density relationship. gravity and without the help of external agency. On the other hand effusion is a special case of diffusion when a gas is allowed to escape through a small orifice or hole.