Question

At a certain temperature, the solubility of n2 gas in water at 3.08 atm is 72.5 mg of n2 gas/100 g water . calculate the solubility of n2 gas in water, at the same temperature, if the partial pressure of n2 gas over the solution is increased from 3.08 atm to 8.00 atm . express your answer numerically to three significant figures.

Answer 1

The solubility of N2 gas in water at the same temperature, when the partial pressure is increased from 3.08 atm to 8.00 atm, is 187.1 mg/100 g water.

Step-by-step explanation:

According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation is given as:

Solution 2 solubility / Solution 1 solubility = Pressure 2 / Pressure 1

Using the given information:

Pressure 1 = 3.08 atm, Pressure 2 = 8.00 atm, Solution 1 solubility = 72.5 mg/100 g water

Solving for Solution 2 solubility:

Solution 2 solubility = (Pressure 2 / Pressure 1) x Solution 1 solubility

= (8.00 atm / 3.08 atm) x 72.5 mg/100 g

= 187.1 mg/100 g water

So, the solubility of N2 gas in water at the same temperature, when the partial pressure is increased from 3.08 atm to 8.00 atm, is 187.1 mg/100 g water.

Answer 2

According to Henry's law, solubility of solution is directly proportional to partial pressure thus,

`(S_(1))/(P_(1))=(S_(2))/(P_(2))`

Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.

Putting the values in equation:

`(0.725)/(3.08)=(S_(2))/(8)`

On rearranging,

`S_(2)=(0.725* 8)/(3.08)=1.88`

Therefore, solubility will be 1.88 mg of
`N_(2)` gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.