
What is the ratio predicted from Graham’s law for rates of diffusion for \[N{H_3}/HCl\]?
Answer
524.1k+ views
Hint: Graham's Law states that a gas's rate of diffusion or effusion is inversely proportional to its molecular weight squared. To put it another way, at constant pressure and temperature, molecules or atoms with lower molecular masses can effuse faster than molecules or atoms with higher molecular masses.
Complete answer:
We use the following equation,
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \dfrac{{\sqrt {{M_2}} }}{{\sqrt {{M_1}} }}\]
Where,
\[Rat{e_1} = \] The rate of effusion for the first gas
\[Rat{e_2} = \] The rate of effusion for the second gas
\[{M_1} = \]Molar mass of gas 1
\[{M_2} = \]Molar mass of gas 2
We may infer from Graham's law that the rate of diffusion of a gas is inversely proportional to the square root of the gas's molecular mass.
\[rat{e_1} \propto \dfrac{1}{{\sqrt {{m_1}} }}\] and \[rat{e_2} \propto \dfrac{1}{{\sqrt {{m_2}} }}\]
That is,
\[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\sqrt {{m_2}} }}{{\sqrt {{m_1}} }}\]
We know that the molar mass of \[HCl\] is \[36.458g{(mol)^{ - 1}}\] and the molar mass of \[N{H_3}\] is \[17.031g{(mol)^{ - 1}}\].
By substituting the values to the equation, we get
\[\dfrac{{{r_{N{H_3}}}}}{{{r_{HCl}}}} = \dfrac{{\sqrt {36.458} }}{{\sqrt {17.031} }}\]
\[ = 1.46\]
This shows that ammonia gas diffuses at a much slower rate of 1.46 than hydrogen chloride gas.
Additional Information:
Remember that from the previous statement we know that a gas's rate of diffusion or effusion is inversely proportional to its molecular weight squared. If one gas has four times the molecular weight of another, it can diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate. Gases that are heavier disperse more slowly.
Note:
The molar mass is proportional to the mass density at the same temperature and pressure. As a result, the square roots of the mass densities of various gases are inversely proportional to their diffusion rates.
Complete answer:
We use the following equation,
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \dfrac{{\sqrt {{M_2}} }}{{\sqrt {{M_1}} }}\]
Where,
\[Rat{e_1} = \] The rate of effusion for the first gas
\[Rat{e_2} = \] The rate of effusion for the second gas
\[{M_1} = \]Molar mass of gas 1
\[{M_2} = \]Molar mass of gas 2
We may infer from Graham's law that the rate of diffusion of a gas is inversely proportional to the square root of the gas's molecular mass.
\[rat{e_1} \propto \dfrac{1}{{\sqrt {{m_1}} }}\] and \[rat{e_2} \propto \dfrac{1}{{\sqrt {{m_2}} }}\]
That is,
\[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\sqrt {{m_2}} }}{{\sqrt {{m_1}} }}\]
We know that the molar mass of \[HCl\] is \[36.458g{(mol)^{ - 1}}\] and the molar mass of \[N{H_3}\] is \[17.031g{(mol)^{ - 1}}\].
By substituting the values to the equation, we get
\[\dfrac{{{r_{N{H_3}}}}}{{{r_{HCl}}}} = \dfrac{{\sqrt {36.458} }}{{\sqrt {17.031} }}\]
\[ = 1.46\]
This shows that ammonia gas diffuses at a much slower rate of 1.46 than hydrogen chloride gas.
Additional Information:
Remember that from the previous statement we know that a gas's rate of diffusion or effusion is inversely proportional to its molecular weight squared. If one gas has four times the molecular weight of another, it can diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate. Gases that are heavier disperse more slowly.
Note:
The molar mass is proportional to the mass density at the same temperature and pressure. As a result, the square roots of the mass densities of various gases are inversely proportional to their diffusion rates.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

