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Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at 30 ∘C is 7.27×10−3M, what is the Henry's law constant for radon in water at this temperature?

Respuesta :

superman1987

Answer:

K = 137.55 atm/M.

Explanation:

  • The relationship between gas pressure and the  concentration of dissolved gas is given by  Henry’s law:

P = (K)(C)

where P is the partial pressure of the gaseous  solute above the solution (P = 1.0 atm).

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas (C = 7.27 x 10⁻³ M).

∴ K = P/C = (1.0 atm)/(7.27 x 10⁻³ M) = 137.55 atm/M.

mahima6824soni

Henry's law constant for radon in water at this temperature, K is 137.55 atm/M.

What is Henry's law constant?

Henry's law constant states that the amount of dissolved gas in any liquid is proportional to the partial pressure above liquid.

[tex]c = kp[/tex]

c = concentration of dissolved gas (7.27 x 10⁻³ M )

k = Henry' law constant

p = partial pressure

Therefore,

[tex]\bold{k= \dfrac{c}{p}= \dfrac{1.0\; atm}{ 7.27\times10^-^3} = 137.55 atm/M }[/tex]

Thus, 137.55 atm/M is Henry's law constant for radon in water at this temperature.

Learn more about Henry law constant, here:

https://brainly.com/question/6195790

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