Question

hello

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(i) Gas (A) is more soluble in water than Gas (B) at the same temperature. Which one of the two gases will have the higher value of K H (Henry's constant) and why?


(ii) In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?

Answer 1

$\underline{hello dear}$

here is your ans

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(i) According to Henry's law, the solubility of a gas is inversely related to the Henry's constant (K H) for that gas. Hence, gas (B), being less soluble, would have a higher K H value.

(ii) A maximum boiling azeotrope shows negative deviation from the Raoult's law.
so it shows negative deviation .

hope it hlps

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Answer 2

Hey dear,

● Answers -
(i) Gas (B)
(ii) -ve deviation

● Explanation -
(i) Problem I -
- According to Henry's law, Henry's constant of the gas is inversely proportional to solubility of that gas.
∴ S ∝ 1/kH
- As solubility of gas (B) in water is less, its kH value must be high.

(ii) Problem II -
- In a non-ideal soln, formation of maximum boiling azetropes show negative deviation form Raoult's law.

Hope this helps you...