Question

Two gases $A$ and $B$ having the same volume diffuse through a porous partition in $20$ and $10$ seconds respectively , the molecular mass of $A$ is $49amu$ .Molecular mass of $B$ will be
A. $25amu$
B. $50amu$
C. $12.25amu$
D. $6.50amu$

Answer 1

Hint : The molecular mass of a compound is defined as the sum of the atomic masses of the atoms that form the compound. Graham's law is a gas law that relates a molar mass of a gas to its rate of diffusion or effusion. The process of slowly mixing two gases together is known as diffusion. When a gas is allowed to escape its container through a small opening, this is known as effusion.
$\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}}$
$Rat{e_1} =$ is the rate of effusion of the first gas
$Rat{e_2} =$ is the rate of effusion for the second gas
${M_1} =$ molar mass of gas of the first gas
${M_2} =$ molar mass of gas of the second gas.

Complete Step By Step Answer:
Rate of Gas $A = 20$
Rate of Gas $B = 10$
Molecular mass of $A = 49amu$
To find = Molecular mass of $B$
Substituting the values in the above equation,
$\dfrac{{10}}{{20}} = \sqrt {\dfrac{{{M_B}}}{{49}}}$
${\left[ {\dfrac{{10}}{{20}}} \right]^2} = \dfrac{{{M_B}}}{{49}}$
$\dfrac{{100}}{{400}} = \dfrac{{{M_B}}}{{49}}$
${M_B} = \dfrac{{49 \times 100}}{{400}}$
On solving the above equation, we get,
${M_B} = 12.25amu$
Hence, the correct option is C. $12.25amu$ .

Note :
Thomas Graham studied the effusion process and discovered an essential characteristic that lighter gas molecules travel quicker than heavier gas molecules. Graham's Law states that at constant pressure and temperature, lower molecular mass molecules or atoms will effuse faster than greater molecular mass molecules or atoms. Thomas even discovered the pace at which they disperse through diffusion. In other words, the rate of gas effusion is inversely proportional to the square root of the molecular mass of the gas.