Question

Two gases \[A\] and \[B\] having the same volume diffuse through a porous partition in \[20\] and \[10\] sec respectively. The molecular mass of \[A\] is \[49{\text{ }}u\]. The molecular mass of \[B\] will be:
A) \[12.25{\text{ }}u\]
B) \[6.50{\text{ }}u\]
C) \[25.00{\text{ }}u\]
D) \[50.00{\text{ }}u\]

Answer 1

Hint: Graham’s law of diffusion can be defined as, at constant temperature and pressure, gaseous molecules or gaseous atoms having lower molecular mass move faster than molecules having higher molecular mass. The gas molecules are classified as two types based on the density of the gas. If the gas molecules have more density than heavier gas molecules and if the gas molecules have lower density than they are called lighter gas molecules. The lighter gas molecules move faster than heavier gas molecules. The movement of gas molecules through other gas molecules is known as diffusion.
Formula used:
Graham’s law of diffusion can be defined as follows:
Graham’s law is that the rate of diffusion of the gas is inversely proportional to the square root of the molecular mass of that gas.
Rate of diffusion of the gas \[{{\propto }}\dfrac{{\text{1}}}{{\sqrt {\text{M}} }}\]
M is the symbol used for molecular mass. Molecular mass can be defined as the sum of atomic masses of all the elements present in the given compound.
The example for this law is
\[\dfrac{{{\text{Rat}}{{\text{e}}_{\text{1}}}}}{{{\text{Rat}}{{\text{e}}_{\text{2}}}}} = \dfrac{{\sqrt {{{\text{M}}_{\text{2}}}} }}{{{{\sqrt {\text{M}} }_1}}}\]
Here,
\[{{\text{M}}_1} = \] Molecular mass of gas\[1\].
\[{{\text{M}}_2} = \] Molecular mass of gas\[2\].
Rate is equal to the ratio between the volume and time.
\[Rate{\text{ }} = {\text{ }}\dfrac{{volume}}{{time}}\]

Complete answer:
The given data is below,
Two gases are given:
\[A\]and \[B\] have the same volume diffused through a porous partition in \[20\] and \[10\] sec respectively.
The molecular mass of \[A\] is \[49{\text{ }}u\].
The molecular mass of \[B\] will be calculated,
The mathematical form of Graham’s law of diffusion is
\[\dfrac{{{\text{Rat}}{{\text{e}}_{\text{A}}}}}{{{\text{Rat}}{{\text{e}}_{\text{B}}}}} = \dfrac{{\sqrt {{{\text{M}}_{\text{B}}}} }}{{{{\sqrt {\text{M}} }_A}}}\]
\[\dfrac{{{\text{Rat}}{{\text{e}}_{\text{A}}}}}{{{\text{Rat}}{{\text{e}}_{\text{B}}}}} = \dfrac{{\sqrt {{{\text{M}}_{\text{B}}}} }}{{\sqrt {49{\text{ }}u} }}\]
We applying new formula is
\[Rate{\text{ }} = {\text{ }}\dfrac{{volume}}{{time}}\]
But the volume of both gases is the same. The formula will change as
 \[\dfrac{{{\text{tim}}{{\text{e}}_{\text{B}}}}}{{{\text{tim}}{{\text{e}}_{\text{A}}}}} = \dfrac{{\sqrt {{{\text{M}}_{\text{B}}}} }}{{\sqrt {49{\text{ }}u} }}\]
\[\dfrac{{10}}{{{\text{20}}}} = \dfrac{{\sqrt {{{\text{M}}_{\text{B}}}} }}{{\sqrt {49{\text{ }}u} }}\]
\[\dfrac{1}{2} = \dfrac{{\sqrt {{{\text{M}}_{\text{B}}}} }}{{\sqrt {49{\text{ }}u} }}\]
Squaring on both sides, we get the answer is
\[\dfrac{1}{4} = \dfrac{{{{\text{M}}_{\text{B}}}}}{{49{\text{ }}u}}\]
\[{{\text{M}}_{\text{B}}} = \dfrac{{49{\text{ }}u}}{4}\]
\[{{\text{M}}_{\text{B}}} = 12.25\;u\]
From the above calculation, we can say that the molecular mass of \[A\] is \[49{\text{ }}u\]. The molecular mass of \[B\] will be \[12.25\;u\].

Hence, the correct answer is option A.

Note:
The moles are one of the main units in chemistry. The moles of the molecule depend on the mass of the molecule and molecular mass of the molecule. Chemical reactions are measured by moles only. The chemical properties of the gas depend on the mixture of gases. All the gases are not helpful for the human body and some gases are poison for the human body and lead to death also.