Question

A mixture of $S{{O}_{2}}$, and ${{O}_{2}}$ in the molar ratio 16 : 1 is diffused through a pinhole for successive effusions three times to give a molar ratio 1 : 1 of diffused mixture. Which one is not correct if diffusion is made at the same P and T in each operation?
$I.$ Eight operations are needed to get 1 : 1 molar ratio.
$II.$ Rate of diffusion for$S{{O}_{2}}$:${{O}_{2}}$ after eight operations is 0.707.
$III.$ Six operations are needed to get 2 : 1 molar ratio for $S{{O}_{2}}$ and ${{O}_{2}}$ in diffusion mixture.
$IV.$ Rate of diffusion for $S{{O}_{2}}$ and ${{O}_{2}}$ after six operations is 2.41.
A. $I,II,III$
B. $II,III$
C. $I,III$
D. $IV$

Answer 1

Hint: Diffusion id the movement of molecules across concentration gradients. While, effusion is considered to be the movement of molecules between two containers.

Formula used: Rate of diffusion $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}$ where, M is the molar masses of 2 gases.

Complete step by step answer: We have been given two gases $S{{O}_{2}}$ and ${{O}_{2}}$, the initial molar ratio for them is 16 : 1 respectively. After successive effusions, the molar ratio become 1 : 1 times the diffusion mixture. Now we will see, for these gases which of the statements are correct. For this we will find out the rate of diffusion, keeping the molar ratios of initial and final concentration.
Now consider, $X\,\log \,{{f}_{1}}=\,\log \,\left[ \dfrac{{{n}^{'}}_{S{{O}_{2}}}}{{{n}^{'}}_{{{O}_{2}}}}\times \dfrac{{{n}_{{{O}_{2}}}}}{{{n}_{S{{O}_{2}}}}} \right]$ where ${{n}^{'}}$ is the initial molar ratios, and n is the final molar ratios.
Now, through formula of rate of diffusion, $X\,\log \,\sqrt{\dfrac{{{M}_{{{O}_{2}}}}}{{{M}_{S{{O}_{2}}}}}}=\,\log \,\left[ \dfrac{{{n}^{'}}_{S{{O}_{2}}}}{{{n}^{'}}_{{{O}_{2}}}}\times \dfrac{{{n}_{{{O}_{2}}}}}{{{n}_{S{{O}_{2}}}}} \right]$
Putting the molar masses of gases and ${{n}^{'}}$to be 1 : 1 and n to be 1 : 16, we have,
$X\,\log \,\sqrt{\dfrac{32}{64}}=\,\log \left[ \dfrac{1}{1}\times \dfrac{1}{16} \right]$, we have, X= 8
So, after 8 operation the rate, $\dfrac{{{n}_{1}}}{{{n}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}$ = 0.707.
Now if number of operations, X = 6, then, $6\,\log \,\sqrt{\dfrac{32}{64}}=\,\log \,\left[ \dfrac{{{n}^{'}}_{S{{O}_{2}}}}{{{n}^{'}}_{{{O}_{2}}}}\times \dfrac{{{n}_{{{O}_{2}}}}}{{{n}_{S{{O}_{2}}}}} \right]$,
So, $6\,\log \,\sqrt{\dfrac{32}{64}}=\,\log \,\left[ \dfrac{{{n}^{'}}_{S{{O}_{2}}}}{{{n}^{'}}_{{{O}_{2}}}}\times \dfrac{1}{16} \right]$,
Therefore, $\dfrac{{{n}^{'}}_{S{{O}_{2}}}}{{{n}^{'}}_{{{O}_{2}}}}$ = 2 : 1
So, the rate of diffusion is 0.707 ion each operation.

Hence, the statements, $I,II,III$ are correct, and option A is the right option.

Note: X is considered to be the number of operations, it can also be calculated by, ${{({{f}_{1}})}^{x}}=\dfrac{{{n}^{'}}_{S{{O}_{2}}}}{{{n}^{'}}_{{{O}_{2}}}}\times \dfrac{{{n}_{{{O}_{2}}}}}{{{n}_{S{{O}_{2}}}}}$ , where ${{n}^{'}}$ is the initial molar ratios, and n is the final molar ratios.