Question

Calculate the mass of nitrogen dissolved at room temperature in an 89.0 LL home aquarium. Assume a total pressure of 1.0 atmatm and a mole fraction for nitrogen of 0.78.

Answer 1

Answer : The mass of nitrogen dissolved is, 79.4 grams

Explanation :

The Raoult's law for liquid phase is:

`p_(N_2)=x_(N_2)* p_T`

where,

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

`p_T` = total pressure = 1.0 atm

`x_(N_2)` = mole fraction of nitrogen = 0.78

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

`p_(N_2)=0.78* 1.0atm`

`p_(N_2)=0.78atm`

Now we have to calculate the mass of nitrogen.

Using ideal gas equation:

`PV=nRT\\\\PV=(w)/(M)RT`

where,

P = pressure of gas = 0.78 atm

V = volume of gas = 89.0 L

T = temperature of gas =
`25^oC=273+25=298K`

R = gas constant = 0.0821 L.atm/mole.K

w = mass of gas = ?

M = molar mass of nitrogen gas = 28 g/mole

Now put all the given values in the ideal gas equation, we get:

`(0.78atm)* (89.0L)=(w)/(28g/mole)* (0.0821L.atm/mole.K)* (298K)`

`w=79.4g`

Therefore, the mass of nitrogen dissolved is, 79.4 grams