Question

Henry’s law constant for the molality of methane in benzene at 298 K is $4.27\times {{10}^{5}}$mm Hg. Calculate the solubility of methane in benzene at 298 K under 760mm Hg.

Answer 1

By using Henry’s law we can calculate the solubility of the methane in benzene.
$P={{K}_{H}}x$
Where P = Partial pressure
${{K}_{H}}$= Henry’s law constant
x = mole fraction or solubility of the methane gas

Complete answer:
In the question it is given that Henry’s law constant ${{K}_{H}}=4.27\times {{10}^{5}}mm\text{ }Hg$at 298 K.
The partial pressure P = 760 mm
We can substitute the above values in Henry’s law to find the solubility of methane gas in benzene.

& P={{K}_{H}}x \\\ & x=\dfrac{P}{{{K}_{H}}} \\\ & x=\dfrac{760mm}{4.27\times {{10}^{5}}mm} \\\ & x=1.78\times {{10}^{-5}} \\\ \end{aligned}$$ Therefore the solubility of methane in benzene at 298 K under 760 mm Hg is$$1.78\times {{10}^{-5}}$$. **Note:** Henry's law is a gas law, it states that the amount of gas dissolved in a liquid or solvent is directly proportional to its partial pressure of the liquid. The proportionality constant is called Henry's law constant and it is denoted with a symbol$${{K}_{H}}$$. Henry’s law is only applicable to the molecules of the system where they are in equilibrium. Under extremely high pressure Henry’s law is not applicable. If gas and the solvent or liquid chemical react to each other then also Henry’s law is not applicable. The factors affecting Henry’s law constant are The nature of the gas, nature of the solvent and temperature and pressure of the gas. Therefore different gases have different values of Henry’s law constant in different solvents.