Question

Calculate the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm (a typical value at high altitude). Atmospheric Gas Mole Fraction kH mol/(L*atm) N2 7.81 x 10-1 6.70 x 10-4 O2 2.10 x 10-1 1.30 x 10-3 Ar 9.34 x 10-3 1.40 x 10-3 CO2 3.33 x 10-4 3.50 x 10-2 CH4 2.00 x 10-6 1.40 x 10-3 H2 5.00 x 10-7 7.80 x 10-4

Answer 1

To calculate the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm, we can use Henry's Law. Multiply the mole fraction of nitrogen by the atmospheric pressure to calculate the partial pressure. Then, multiply the Henry's Law constant for nitrogen by the partial pressure to find the solubility of nitrogen in water.

Step-by-step explanation:

To calculate the solubility of nitrogen in water at a given atmospheric pressure, we can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

The equation for Henry's Law is:

C = kH × P

where C is the concentration of the gas in the liquid, kH is the Henry's Law constant, and P is the partial pressure of the gas above the liquid.

In this case, we need to use the mole fraction of nitrogen and the Henry's Law constant for nitrogen to calculate the solubility. The mole fraction of nitrogen is 7.81 x 10^-1 and the Henry's Law constant for nitrogen is 6.70 x 10^-4 mol/(L×atm).

1. Convert the mole fraction of nitrogen to partial pressure using the given atmospheric pressure of 0.480 atm. Multiply the mole fraction by the atmospheric pressure: 7.81 x 10^-1 ×0.480 atm = 3.7572 x 10^-1 atm.
2. Use the Henry's Law constant for nitrogen and the partial pressure of nitrogen calculated in step 1 to find the solubility of nitrogen in water. Multiply the Henry's Law constant by the partial pressure: 6.70 x 10^-4 mol/(L×atm) ×3.7572 x 10^-1 atm = 2.5108 x 10^-4 mol/L.

Therefore, the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm is 2.5108 x 10^-4 mol/L.

Answer 2

Answer : The solubility of nitrogen in water at an atmospheric pressure will be,
`2.5125* 10^(-4)mole/L`

Explanation :

First we have to calculate the partial pressure of nitrogen.

Formula used :

`p_(N_2)=X_(N_2)* P_(atm)`

where,

`p_(N_2)` = partial pressure of nitrogen = ?

`X_(N_2)` = mole fraction of nitrogen =
`7.81* 10^(-1)`

`p_(atm)` = atmospheric pressure = 0.480 atm

Now put all the given values in the above formula, we get :

`p_(N_2)=7.81* 10^(-1)* 0.480 atm`

`p_(N_2)=0.375atm`

Now we have to calculate the solubility of nitrogen in water.

Formula used :

`s_(N_2)=p_(N_2)* K_H`

where,

`p_(N_2)` = partial pressure of nitrogen = 0.375 atm

`s_(N_2)` = solubility of nitrogen in water = ?

`K_H` = Henry's constant =
`6.70* 10^(-4)mole/L.atm`

Now put all the given values in the above formula, we get :

`s_(N_2)=0.375atm* 6.70* 10^(-4)mole/L.atm`

`s_(N_2)=2.5125* 10^(-4)mole/L`

Therefore, the solubility of nitrogen in water at an atmospheric pressure will be,
`2.5125* 10^(-4)mole/L`