Question

The relative rate of diffusion of a gas $\left( {mol.wt = 98} \right)$ as compared to hydrogen will be:
$A.$ $\dfrac{1}{7}$
$B.$ $\dfrac{1}{5}$
$C.$ $\dfrac{1}{4}$
$D.$ $1$

Answer 1

According to the Graham’s law of diffusion the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Let us consider the rate of diffusion $\left( r \right)$ and the molar mass $\left( M \right)$ . so, mathematically this law can be represented as; $r \propto \dfrac{1}{M}$ .

Complete solution:
In the given question it is given that the molecular mass of an unknown gas is $98$ and the molar mass of hydrogen gas is $2.$ We have to calculate the rate of diffusion of unknown gas, we can calculate the rate of diffusion of gas as compared to hydrogen gas by using Graham’s law of diffusion.
Let the rate diffusion of a gas be $\left( {{r_1}} \right)$ and the rate of diffusion of hydrogen gas be $\left( {{r_2}} \right)$ . According to the Graham’s law, the rate of diffusion of hydrogen gas is${r_1} \propto \dfrac{1}{{\sqrt {{M_H}} }}$ and the rate of diffusion of gas is ${r_2} \propto \dfrac{1}{{\sqrt {{M_g}} }}$ where ${M_H}$ and ${M_g}$ is the molar mass of hydrogen and gas respectively. On comparing the rate of diffusion of a gas with hydrogen gas, we get the relation,
$\dfrac{{{r_2}}}{{{r_1}}} = \dfrac{{\dfrac{{\dfrac{1}{{\sqrt {{M_g}} }}}}{1}}}{{\sqrt {{M_H}} }}$
$\Rightarrow$ $\dfrac{{{r_2}}}{{{r_1}}} = \sqrt {\dfrac{{{M_H}}}{{{M_g}}}}$
By putting the value of given quantity, we get
$\dfrac{{{r_2}}}{{{r_1}}} = \sqrt {\dfrac{2}{{98}}}$
$= \dfrac{1}{7}$
Hence, the correct option is $A.$

Note: It is to be noted that when gases are diffusing through other gases, their rate of diffusion can be defined by Graham’s law; it state that the rate of diffusion is inversely proportional to the square root of its molar mass at identical pressure and temperature, In other word we can say the smaller mass of the gas, the more rapidly it will diffuse.