Question

$180$ ml of hydrocarbon having the molecular weight $16$ diffuses in $1.5$min. Under similar conditions time taken by $120$ ml of $S{{O}_{2}}$ to diffuse is:
A. $2$min
B. $1.5$ min
C. $1$ min
D. $1.75$min

Answer 1

This question is based on the concept of graham’s law of diffusion. This law relates the rate of effusion to the molar masses of the compounds. If you are aware about the mathematical expression of this law, then you can easily find out the solution to the given problem.

Complete step-by-step solution: Let us firstly understand the process of diffusion.
Diffusion is the process of intermixing of two or more gases irrespective of density, gravity and without the help of external agency.
Rates of diffusion is elaborated on the basis of Graham’s law which states that under similar condition of pressure and temperature the rates of diffusion are inversely proportional to the square roots of their molecular masses.
Let ${{r}_{1}}\And {{r}_{2}}$are the rates of two gases of molar masses ${{M}_{1}}\And {{M}_{2}}$
Therefore, by graham’s law $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}$
Now, we know that rate (r) of diffusion of gas can be written as $r=\dfrac{Volume\left( V \right)}{time\left( t \right)}$
Substituting, the value r in the above expression of graham's law we have
$\begin{aligned} & \Rightarrow \dfrac{\dfrac{{{V}_{1}}}{{{t}_{1}}}}{\dfrac{{{V}_{2}}}{{{t}_{2}}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}} \\\ & \Rightarrow \dfrac{{{V}_{1}}}{{{t}_{1}}}\times \dfrac{{{t}_{2}}}{{{V}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}\,\,\,(equation\,1) \\\ \end{aligned}$
In this question, ${{V}_{1}}$= Volume of hydrocarbon = $180$ml
${{V}_{2}}$= Volume of sulfur dioxide = $120$ml
${{M}_{1}}$= Molar mass of hydrocarbon = 16g/mole
${{M}_{2}}$= Molar mass of $S{{O}_{2}}$= $32+2\left( 16 \right)=32+32=64$g/mole
${{t}_{1}}$= time taken for hydrocarbon to diffuse = $1.5$min
And, we need to calculate ${{t}_{2}}$ which refers to time taken by $S{{O}_{2}}$ to diffuse.
Therefore, substituting the given values in equation 1, we get.
$\begin{aligned} & \Rightarrow \dfrac{180}{1.5}\times \dfrac{{{t}_{2}}}{120}=\sqrt{\dfrac{64}{16}} \\\ & \Rightarrow \dfrac{180}{1.5}\times \dfrac{{{t}_{2}}}{120}=2 \\\ & \Rightarrow {{t}_{2}}=\dfrac{2\times 1.5\times 120}{180}=2\min \\\ & \therefore {{t}_{2}}=2\min \\\ \end{aligned}$
Therefore, 2 minutes are required for sulfur dioxide to diffuse.

Hence, the correct option is A. $2$ min.

Note: It may be noted that further rate of diffusion $\left( r \right)\propto \dfrac{P}{\sqrt{M}}$ at constant pressure.
Hence, $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{{{P}_{1}}}{{{P}_{2}}}\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}$. The grahams relation can also be written as
$\dfrac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}\,$where ${{n}_{1}}\,\And {{n}_{2}}$are moles of gas1 and gas 2.