One mole of nitrogen gas at ${0.8atm}$ takes ${38 seconds}$ to diffuse through a pinhole whereas one mole of an unknown compound of xenon with fluorine at ${1.6 atm}$ takes ${57 seconds}$ to diffuse through the same hole.
Calculate the molecular formula of the compound.
Answer
644.4k+ views
Hint: Graham's law of diffusion states that under similar conditions of temperature and pressure, the rate at which gases diffuse is inversely proportional to the square root of their densities.
Complete answer:
It is given that;
${ t }_{ 1 }$ = ${ 38 seconds }$
${ t }_{ 2 }$ = ${ 57 seconds }$
${ P }_{ 1 }$ = ${ 0.8atm }$
${ P }_{ 2 }$ = ${ 1.6atm }$
${ M }_{ 1 }$ = ${ 28gmol }^{ -1 }$
${ M }_{ 2 }$ =?
From Graham’s law of diffusion;
${ t }_{ 2 }{ \div t }_{ 1 }{ =P }_{ 1 }{ \div P }_{ 2 }\sqrt { { M }_{ 2 }{ \div M }_{ 1 } } $........(1)
${ t }_{ 1 }$ = time taken by nitrogen gas
${ t }_{ 2 }$ = time taken by the compound
${ P }_{ 1 }$ = pressure of nitrogen gas
${ P }_{ 2 }$ = pressure of the compound
${ M }_{ 1 }$ = molecular formula of nitrogen gas
${ M }_{ 2 }$ = molecular formula of the compound.
Now, putting the given values in equation (1), we get
${ 57\div 38=0.8\div 16 }\sqrt { { M }_{ 2 }{ \div 28 } } $
Now, squaring both the sides
${ (57\div 38{ ) }^{ 2 }=1\div 4 }{ \times M }_{ 2 }{ \div 28 }$
${ M }_{ 2 }{ =(57 }^{ 2 }{ \times 28\times 4\div { 38 }^{ 2 } }$
The molecular weight of the compound = ${ 252 }$
The compound contains Xe and F , so Xe +nF = ${ 252 }$
The molecular mass of Xe = ${ 131 }$
The molecular mass of F = ${ 19 }$
Hence, ${ 131+ n(19) = 252 }$
We get, n =${ 6 }$
So, the molecular formula of the compound is ${ XeF }_{ 6 }$.
Additional Information:
Graham's law is generally precise for molecular effusion which includes the movement of each gas at a time through a hole. It is just approximate for the diffusion of one gas in another or in air, as these procedures involve the movement of more than one gas.
Note: The possibility to make a mistake is that Graham’s law is equal to the square root of the molar mass of gas 2 by the molar mass of gas 1.
Complete answer:
It is given that;
${ t }_{ 1 }$ = ${ 38 seconds }$
${ t }_{ 2 }$ = ${ 57 seconds }$
${ P }_{ 1 }$ = ${ 0.8atm }$
${ P }_{ 2 }$ = ${ 1.6atm }$
${ M }_{ 1 }$ = ${ 28gmol }^{ -1 }$
${ M }_{ 2 }$ =?
From Graham’s law of diffusion;
${ t }_{ 2 }{ \div t }_{ 1 }{ =P }_{ 1 }{ \div P }_{ 2 }\sqrt { { M }_{ 2 }{ \div M }_{ 1 } } $........(1)
${ t }_{ 1 }$ = time taken by nitrogen gas
${ t }_{ 2 }$ = time taken by the compound
${ P }_{ 1 }$ = pressure of nitrogen gas
${ P }_{ 2 }$ = pressure of the compound
${ M }_{ 1 }$ = molecular formula of nitrogen gas
${ M }_{ 2 }$ = molecular formula of the compound.
Now, putting the given values in equation (1), we get
${ 57\div 38=0.8\div 16 }\sqrt { { M }_{ 2 }{ \div 28 } } $
Now, squaring both the sides
${ (57\div 38{ ) }^{ 2 }=1\div 4 }{ \times M }_{ 2 }{ \div 28 }$
${ M }_{ 2 }{ =(57 }^{ 2 }{ \times 28\times 4\div { 38 }^{ 2 } }$
The molecular weight of the compound = ${ 252 }$
The compound contains Xe and F , so Xe +nF = ${ 252 }$
The molecular mass of Xe = ${ 131 }$
The molecular mass of F = ${ 19 }$
Hence, ${ 131+ n(19) = 252 }$
We get, n =${ 6 }$
So, the molecular formula of the compound is ${ XeF }_{ 6 }$.
Additional Information:
Graham's law is generally precise for molecular effusion which includes the movement of each gas at a time through a hole. It is just approximate for the diffusion of one gas in another or in air, as these procedures involve the movement of more than one gas.
Note: The possibility to make a mistake is that Graham’s law is equal to the square root of the molar mass of gas 2 by the molar mass of gas 1.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

The chemical formula of tear gas is A CO Cl 2 B C 10 class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

