First, we need to calculate the concentration of nitrogen in the water using Henry's law:
P = k * X
1.0 atm = (6.1 × 10^-4 M/atm) * X
X = 1.0 atm / (6.1 × 10^-4 M/atm)
X = 1639.34 M
Now we can convert the concentration to moles of nitrogen:
1 L = 1000 cm^3
90.0 L = 90.0 * 1000 cm^3 = 90,000 cm^3
Density of water = 1 g/cm^3
90,000 cm^3 of water weighs 90,000 g
Molar mass of N2 = 14.01 g/mol
Moles of N2 = concentration * volume
Moles of N2 = 1639.34 mol/L * 90,000 L = 147,537.06 mol
Finally, we can calculate the mass of nitrogen:
Mass = moles * molar mass
Mass = 147,537.06 mol * 14.01 g/mol = 2,067,258.78 g
Therefore, the mass of nitrogen dissolved in the 90.0 L home aquarium at room temperature is approximately 2,100,000 g or 2.1 x 10^6 g.
Calculate the mass of nitrogen dissolved at room temperature in an 90.0 L home aquarium. Assume a total pressure of 1.0 atm and a mole fraction for nitrogen of 0.78. The Henry's law constant for nitrogen at 25 °C is 6.1 × 10^-4 M/atm
Express your answer in grams to two significant figures
1 answer