Question

The rate of diffusion of hydrogen is about :
(A). $\dfrac{1}{2}$ that of helium
(B). $1.4$ time that of helium
(C). Twice that of helium
(D). Fore times that of helium.

Answer 1

The ability of a gas to mix spontaneously and to form a homogenous mixture is known as diffusion . The law which deals with this phenomenon is Graham's law .

Complete step by step answer:
Graham's law of diffusion states that under similar conditions of temperature and pressures , the rates of diffusion or effusion of different gases are inversely proportional to the square root of their densities .
Since we know the molecular masses of helium and hydrogen we have to use the formula which has molecular mass in it . The ratio of densities of two gases is equal to the ratio of their vapour densities and also molecular mass is twice of vapour density .
So , according to Graham’s law
$\dfrac{{{r_{{H_2}}}}}{{{r_{He}}}} \propto \sqrt {\dfrac{{{M_{He}}}}{{{M_{{H_2}}}}}}$ -------(1)
Here, ${r_{{H_2}}}$ - rate of diffusion of hydrogen
${r_{{H_e}}}$ - rate of diffusion of helium
${M_{He}}$- molar mass of helium $= 4$
${M_{{H_2}}}$ - molar mass of hydrogen $= 2$
$\dfrac{{{r_{{H_2}}}}}{{{r_{He}}}} = \sqrt {\dfrac{4}{2}} = \sqrt 2 = 1.414$
When we substituted the values in equation $\left( 1 \right)$, we got the answer as $1.414$
that is , the correct option is $\left( B \right)$

Additional information:
Graham’s law is useful in a number of ways , some of them are :
-It helps in the separation of gases having different densities .
-It helps in the separation of isotopes of certain elements, for example Uranium .
- It helps to determine the density or molecular mass of an unknown gas by comparing its rate of diffusion with a known gas .

Note:
Graham’s law is most accurate for molecular effusion which involves the movement of one gas at a time through a hole. In the same condition of temperature and pressure, the molar mass is proportional to the mass density. Therefore, the rates of diffusion of different gasses are inversely proportional to square root of their mass densities.
$r \propto \dfrac{1}{{\sqrt d }}$