Question

if the solubility of a gas is 7.5 g/L at 404 pressure, what is the solubility of the gas when the pressure is 202 kPa?

Answer 1

henry's law :

S₁  / P₁ = x / P₂

7.5 / 404 = x / 202

x = 7.5 * 202 / 404

x = 1515 / 404

x = 3.75 g/L

hope this helps!

Answer 2

Hello!

Data:

C1 (initial concentration) = 7.5 g/L

P1 (initial pressure) = 404 kPa

C2 (final concentration) = ? (g/L)

P2 (final pressure) = 202 kPa

We apply the data to the formula of the solubility of gases in a liquid (Henry's Law), let us see:

$\boxed{ \dfrac{C_1}{P_1} = \dfrac{C_2}{P_2}}$

Solving:  

$\dfrac{7.5}{404} = \dfrac{C_2}{202}$

$404*C_2 = 7.5*202$

$404C_2 = 1515$

$C_2 = \dfrac{1515}{404}$

$\boxed{\boxed{C_2 = 3,75\:g/L}}\end{array}}\qquad\checkmark$

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I Hope this helps, greetings ... Dexteright02!