Answer:
88.48 grams of Nitrogen
Explanation:
To calculate this, you have to use the ideal gas equation
P x V = n x R x T
Where P = pressure (1 atm)
V = Volume (99 L)
n = moles
R = ideal gas constant (0.082 atm x L /(K x mol))
T = Temperature in Kelvin (25 Celsius = 298 K)
(1 atm x 99 L)/0.082atm x L /(K x mol)x298 K = n
4.05 moles = n
This is the total amount of moles in the aquarium, but you are only interested in Nitrogen, and the mole fraction for nitrogen is 0.78, so you have:
4.05 moles x 0.78 = 3.16 moles of Nitrogen
One mol of Nitrogen (N2) has a mass of 28 grams, so in 3.16 moles you have:
3.16 x 28 grams = 88.5 grams of Nitrogen
Explanation:
The given data is as follows.
[tex]P_{total}[/tex] = 1 atm, [tex]X_{NO_{2}}[/tex] = 0.78
Now, we will calculate the partial pressure of given nitrogen gas as follows.
[tex]P_{NO_{2}} = X_{2}P_{total}[/tex]
= [tex]0.78 \times 0.78[/tex]
= 0.78 atm
According to the Henry's law we will calculate the solubility of nitrogen as follow.
[tex]S_{N_{2}} = K_{H}P_{N_{2}}[/tex]
= [tex]6.1 \times 10^{-4} M/atm \times 0.78 atm[/tex]
= [tex]4.758 \times 10^{-4} M[/tex]
Now, for 99.0 L the amount of nitrogen gas dissolved is calculated as follows.
[tex]\frac{99.0 L N_{2} \times 4.758 \times 10^{-4} M \times 28.0134 g N_{2}}{1 L N_{2} \times 1 mol N_{2}}[/tex]
= [tex]13195.49 \times 10^{-4}[/tex]
= 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.
