Question

Given, 100 mL of ${{O}_{2}}$ gas diffuses in 10 sec. 100mL of gas X diffuses in the sec. Gas X and time t can be:
A) ${{H}_{2}}$ , 2.5 sec
B) $S{{O}_{2}}$ , 16 sec
C) CO, 10 sec
D) He, 4 sec

Answer 1

Hint: The rate of diffusion of a gas is inversely proportional to the square root of molecular mass of the gas. This law is known as Graham’s law. Find the answer by using the expression for this law given by \[\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}\]

Complete answer:
We have come across the chapters of chemistry which tells about the basic physical chemistry concept that includes some of the laws given for the reactions in solutions.
We shall see Graham’s law in detail to deduce the correct answer.
- Graham’s law states that ‘The rate of diffusion of a gas is inversely proportional to the square root of molecular mass of the gas’.
- Diffusion is the gradual mixing of gases due to the motion of the component particles of the gas even in the absence of mechanical agitation such as stirring. The resultant mixture is a uniform mixture.
- Graham law of diffusion in general can be stated as the rate of diffusion or effusion of two gases is equal to the square root of the inverse ratio of their molar masses.
- The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. The mathematical representation of the law is:
\[\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}\]
where, ${{r}_{1}}$ and ${{r}_{2}}$ are the rate of effusion for gas 1 and 2 respectively.
${{M}_{1}}$ and ${{M}_{2}}$ are the molar mass of gas 1 and 2 respectively.
As in the question, same amount of gas is diffused, therefore we can directly consider rate of diffusion to be inversely proportional to the time taken, thus the equation can be rewritten as,
\[\dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}\]
Now, checking options,
a) ${{H}_{2}}$, 2.5 sec
Now, substituting the values in the equation, \[\dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}\], we have
$\dfrac{10}{2.5}=\sqrt{\dfrac{32}{2}}$
$\Rightarrow 4=4$
Thus, this is correct answer
b) $S{{O}_{2}}$ , 16 sec
Substituting the values in the same equation we have,
$\dfrac{10}{16}=\sqrt{\dfrac{32}{64}}$
$\Rightarrow 0.625=0.707$
Thus, they are not equal and this option is ruled out.
c) CO, 10 sec
Applying the formula,
$\dfrac{10}{10}=\sqrt{\dfrac{32}{28}}$
$\Rightarrow 1=1.069$
Thus, they are not equal and this option is also ruled out.
d) He, 4 sec
Applying the formula,
$\dfrac{10}{4}=\sqrt{\dfrac{32}{4}}$
$\Rightarrow 2.5=2.82$
Thus, this option is also ruled out.

Hence, (A) is the correct answer.

Note:
A student must note that if the volume of the gases diffused isn't the same, then the rate of diffusion isn’t directly equal to the inverse of time taken. In that case, the rate of diffusion is equal to the amount of gas diffused divided by the time taken.