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
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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.
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