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The relationship between solubility of a gas in liquid at constant temperature and external pressure is stated by which law?
(a) Raoult’s law
(b) Van’t Hoff Boyle’s law
(c) Van’t Hoff Charles law
(d) Henry’s law

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Last updated date: 29th Feb 2024
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Hint: Try to recall that the solubility of a gas in a liquid is inversely proportional to the temperature of gas and also the mass of gas dissolved is proportional to the partial pressure of the gas. Now, by using this you can find the correct option from the given options.

Complete step by step solution:
It is known to you that Henry’s law states that “at constant temperature and external pressure, the solubility of a gas in a liquid is directly proportional to the pressure at which it is dissolved.
Let x be the mole fraction of a gas at a given temperature as a measure of its solubility and p be the partial pressure of gas in equilibrium with solution.
By Henry’s law: \[x\alpha p\] or \[p\alpha x\]
\[ \Rightarrow p = {K_H}x\] or \[x = \dfrac{p}{{{K_H}}}\] , where${K_H}$ = Henry law constant.
Therefore, from above we can conclude that option D is the correct option to the given question.

Additional information:
1) There are some limitations of henry’s law which are as follows:
2) The pressure of gas should not be too high.
3) Also, the temperature should not be too low.
4) The gas should not undergo any chemical reaction with solvent.
5) The gas should not undergo dissociation in solution.

Note: It should be remembered to you that higher the value of Henry's law constant $K_H$ , lower will be its solubility.
Also, you should remember that different gas has different values of henry’s law constant $K_H$ for the same solvent.