
Henry’s law constant of molality of methane in benzene at 298K is $4.27\times {{10}^{5}}mm$Hg. The mole fraction of methane in benzene at 298K under 760mm Hg is:
[A]$1.78\times {{10}^{-3}}$
[B] 17.43
[C] 0.114
[D] 2.814
Answer
511.8k+ views
Hint: Henry’s law gives us the relation between the partial pressure and the solubility of a gas at equilibrium. To solve this problem, we can use the Henry’s law formula which is\[{{P}_{g}}={{K}_{H}}x\]where ‘x’ is the mole fraction, ‘P’ is the pressure and ‘K’ is the proportionality constant.
Complete answer:
The Henry’s law states that “At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.” We can write the Henry’s law symbolically as-
\[C={{K}_{H}}{{P}_{g}}\]
Where, C is the solubility of the gas in the given liquid at a particular temperature.
K is Henry's law proportionality constant.
P is the partial pressure of the gas.
In the given question, the temperature is constant which means we can use the Henry’s law formula to find out the mole fraction of methane in benzene.
We can also write the Henry’s law formula in terms of mole fraction as-
\[{{P}_{g}}={{K}_{H}}x\]
Where the terms have their usual meaning and ‘x’ is the mole fraction of the solute in the solvent.
The Henry’s law constant is given to us in the question and it is $4.27\times {{10}^{5}}mm$ and the pressure is 760mm.
Putting these values in the above equation, we get
\[\begin{align}
& 760=4.27\times {{10}^{5}}mm\text{ }\times \text{ x} \\
& \text{or, x=1}\text{.78}\times \text{1}{{\text{0}}^{-3}} \\
\end{align}\]
The mole fraction of methane in benzene at 298K is$1.78\times {{10}^{-3}}$.
Therefore, the correct answer is option [A] $1.78\times {{10}^{-3}}$.
Note:
We can apply Henry’s law only when the molecules are at equilibrium. It is also inapplicable for gases at high pressure and also if the solute and solvent undergoes a chemical reaction like dissociation or displacement.
Complete answer:
The Henry’s law states that “At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.” We can write the Henry’s law symbolically as-
\[C={{K}_{H}}{{P}_{g}}\]
Where, C is the solubility of the gas in the given liquid at a particular temperature.
K is Henry's law proportionality constant.
P is the partial pressure of the gas.
In the given question, the temperature is constant which means we can use the Henry’s law formula to find out the mole fraction of methane in benzene.
We can also write the Henry’s law formula in terms of mole fraction as-
\[{{P}_{g}}={{K}_{H}}x\]
Where the terms have their usual meaning and ‘x’ is the mole fraction of the solute in the solvent.
The Henry’s law constant is given to us in the question and it is $4.27\times {{10}^{5}}mm$ and the pressure is 760mm.
Putting these values in the above equation, we get
\[\begin{align}
& 760=4.27\times {{10}^{5}}mm\text{ }\times \text{ x} \\
& \text{or, x=1}\text{.78}\times \text{1}{{\text{0}}^{-3}} \\
\end{align}\]
The mole fraction of methane in benzene at 298K is$1.78\times {{10}^{-3}}$.
Therefore, the correct answer is option [A] $1.78\times {{10}^{-3}}$.
Note:
We can apply Henry’s law only when the molecules are at equilibrium. It is also inapplicable for gases at high pressure and also if the solute and solvent undergoes a chemical reaction like dissociation or displacement.
Recently Updated Pages
Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

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

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

Trending doubts
Explain why it is said like that Mock drill is use class 11 social science CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

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

Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

Net gain of ATP in glycolysis a 6 b 2 c 4 d 8 class 11 biology CBSE
