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:

$p_{N_2}=x_{N_2}\times p_T$

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:

$p_{N_2}=0.78\times 1.0atm$

$p_{N_2}=0.78atm$

Now we have to calculate the mass of nitrogen.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT$

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:

$(0.78atm)\times (89.0L)=\frac{w}{28g/mole}\times (0.0821L.atm/mole.K)\times (298K)$

$w=79.4g$

Therefore, the mass of nitrogen dissolved is, 79.4 grams