Question

Calculate the mass of nitrogen dissolved at room temperature in an 99.0 L home aquarium. Assume a total pressure of 1.0 atm and a mole fraction for nitrogen of 0.78. Express your answer using two significant figures.

Answer 1

Step-by-step explanation:

The given data is as follows.

`P_(total)` = 1 atm,
`X_{NO_(2)}` = 0.78

Now, we will calculate the partial pressure of given nitrogen gas as follows.

`P_{NO_(2)} = X_(2)P_(total)`

=
`0.78 * 0.78`

= 0.78 atm

According to the Henry's law we will calculate the solubility of nitrogen as follow.

`S_{N_(2)} = K_(H)P_{N_(2)}`

=
`6.1 * 10^(-4) M/atm * 0.78 atm`

=
`4.758 * 10^(-4) M`

Now, for 99.0 L the amount of nitrogen gas dissolved is calculated as follows.

`(99.0 L N_(2) * 4.758 * 10^(-4) M * 28.0134 g N_(2))/(1 L N_(2) * 1 mol N_(2))`

=
`13195.49 * 10^(-4)`

= 1.32 g

Thus, we can conclude that the mass of nitrogen dissolved at room temperature in an 99.0 L home aquarium is 1.32 g.