Question

Three grams of helium diffuses from a container in 15 minutes. The mass of sulphur dioxide diffusing from the same container over the same time interval will be
(A) 3 g
(B) 6 g
(C) 9 g
(D) 12 g

Answer 1

Hint The given question is all about the Graham’s law of diffusion or effusion. This law states that when the temperature and pressure are constant then the atoms with high molar mass will effuse slower than that of the atoms with low molar mass.

Complete step by step solution:
Let us discuss the basics required to solve the given illustration;
Diffusion- It is the process through which the particles of a gas will move into another gas causing disorder in the entire system.
The change in diffusing molecules over the period of time is known as the rate of diffusion and can be described as;
Rate of diffusion $ \propto \dfrac{1}{\sqrt{density}}\propto \dfrac{1}{\sqrt{M}} $ where, M is the molar mass of gas.
Graham’s law-
It states that the square root of the molar mass is inversely proportional to the rate of diffusion or effusion, in general. Thus, it can be described as;
\[\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}\]
Illustration-
Given that the time required will be the same for both helium and sulphur dioxide. Thus,
\[\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}\]
By solving we get,
 $ \begin{align}
\dfrac{{{r}_{He}}}{{{r}_{S{{O}_{2}}}}}=\dfrac{{{n}_{He}}}{{{n}_{S{{O}_{2}}}}}=\sqrt{\dfrac{{{M}_{S{{O}_{2}}}}}{{{M}_{He}}}} \\
& \dfrac{{}^{3g}/{}_{4g/mol}}{{}^{x}/{}_{64g/mol}}=\sqrt{\dfrac{64g/mol}{4g/mol}} \\
& x=12g \\
\end{align} $

Hence, option (D) is correct.

Note: Do note that the effusion and diffusion are interchangeably used in the definitions and laws but both have completely different meanings.
Effusion is the phenomenon of particles moving from a tiny opening in the vacuum of space or an open container. The process in which the particles escapes from a closed space with time is known as the rate of effusion and can be described as;
Rate of effusion $ \propto \dfrac{1}{\sqrt{density}}\propto \dfrac{1}{\sqrt{M}} $ where, M is the molar mass of gas.