Question

At the STP, 16ml of ${O_2}$ diffused through a porous partition in t seconds. What volume of $C{O_2}$ will diffuse the same time and under the same condition?
(A) 13.65
(B) 10.5ml
(C) 20.2ml
(D) 224.8ml

Answer 1

To solve this question we should have an idea of the Graham's law of diffusion. The whole question revolves around Graham's law of diffusion as it gives a relationship between diffusion and volume. Thus, it is very important to understand the concept behind this law.

Complete step by step answer:
Graham's law of diffusion states that "At a constant temperature and constant pressure gradient the rates of diffusion of different gases are inversely proportional to the square root of their densities".
${r_1} \propto \dfrac{1}{{\sqrt {{d_1}} }}$, ${r_2} \propto \dfrac{1}{{\sqrt {{d_2}} }}$
Where, ${r_1}$ = rate of diffusion of gas 1
${r_2}$= rate of diffusion of gas 2
${d_1}$= density of gas 1
${d_2}$= density of gas 2${d_2}$
Thus, $\dfrac{{{r_1}}}{{{r_2}}} = \sqrt {\dfrac{{{d_2}}}{{{d_1}}}}$
The law is very useful for calculating molecular weight, volume, density etc.
Let solve the problem,
Given: Volume of ${O_2}$= 16ml
Time of flow for both gases = t
Molecular weight (${m_1}$) of ${O_2}$ = 32g
Molecular weight (${m_2}$) of $C{O_2}$= 44g
Let the volume of $C{O_2}$= V
The rate of diffusion is given by:
Rate of diffusion = $\dfrac{{volume}}{{time}}$
Let the rate of diffusion of ${O_2}$ be ${r_1}$and the rate of diffusion of $C{O_2}$ be ${r_2}$.
Rate of diffusion of ${O_2}$ : ${r_1}$= $\dfrac{{16}}{t}$
Rate of diffusion of $C{O_2}$: ${r_2}$= $\dfrac{V}{t}$
According to graham's law:
$\Rightarrow \dfrac{{{r_1}}}{{{r_2}}} = \sqrt {\dfrac{{{m_2}}}{{{m_1}}}}$
$\Rightarrow \dfrac{{16 \times t}}{{V \times t}} = \sqrt {\dfrac{{44}}{{32}}}$
$\Rightarrow V = 13.64g$

Thus, option A is 13.64g.

Note: According to Graham's law of diffusion:
$\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{{m_1}}}{{{m_2}}} = \sqrt {\dfrac{{{d_2}}}{{{d_1}}}} = \dfrac{{{t_2}}}{{{t_1}}}$
Where, ${d_1}$ and ${d_2}$ are densities of gas 1 and gas 2.
Thus, this law is useful for calculating the rate of diffusion, volume, density etc. In this question the time of flow of both${O_2}$ and $C{O_2}$ gases are the same, that is ' t'.