Calculate the mass of nitrogen dissolved at room temperature in an 93.0 LL home aquarium. Assume a total pressure of 1.0 atmatm and a mole fraction for nitrogen of 0.78.

84.6g of nitrogen is the mass dissolved in a home aquarium

Explanation:

Using combined gas law:

PV = nRT

We can solve for the mass of nitrogen solving the total moles of the aquarium and with the mole fraction as follows:

P = 1.0atm

V = 93.0L

R = 0.082atmL/molK

T = 20°C + 273.15 = 293.15K Assuming room temperature as 20°C

PV / RT = n

1.0atm*93.0L / 0.082atmL/molK*293.15K

3.87 = moles of gas

As mole fraction of nitrogen in the air is 0.78, moles of nitrogen are:

3.87 * 0.78 = 3.02moles of nitrogen

As molar mass of nitrogen gas, N₂, is 28g/mol:

3.02 mol N₂ * (28g / mol) =

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explanation:

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the answer is c i believe