Question

The solubility of a gas is 2.0 g/L at 50.0 kPa of pres-

sure. How much gas will dissolve in 1 L at a pressure
of 10.0 kPa?
our of

Answer 1

Answer: 0.4 g/L

Step-by-step explanation:

S2 = S1 x P2 / P1

S2= 2 g/L x 10 kPa / 50 kPa

CROSS OUT

S2= 2g/L x 10 / 50

S2= 20 g/L/50

S2= .4 g/L

Answer 2

That means, under 10.0kPa of pressure, 0.4g of gas can be dissolved in 1L

Step-by-step explanation:

Based on Henry's law, the solubility of the gas is directly proportional to the pressure. The equation is:

P1S2 = P2S1

Where P is pressure and S solubility of 1, initial state and 2, final state of the gas.

Replacing:

P1 = 50.0kPa

S1 = 2.0g/L

P2 = 10.0kPa

S2 = ??

50.0kPa*S2 = 10.0kPa*2.0g/L

S2 = 0.4g/L

That means, under 10.0kPa of pressure, 0.4g of gas can be dissolved in 1L