
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
483k+ views
Hint: 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'.
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'.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What is the modal class for the following table given class 11 maths CBSE

How do I convert ms to kmh Give an example class 11 physics CBSE

Give an example of a solid solution in which the solute class 11 chemistry CBSE
