The solubility of nitrogen in water at 25 °C= 4.88 x 10⁻⁴ mol/L
<h3>Further explanation</h3>
Given
78% Nitrogen by volume
Required
The solubility of nitrogen in water
Solution
Henry's Law states that the solubility of a gas is proportional to its partial pressure
Can be formulated
S = kH. P.
S = gas solubility, mol / L
kH = Henry constant, mol / L.atm
P = partial gas pressure
In the standard 25 C state, the air pressure is considered to be 1 atm, so the partial pressure of N₂ -nitrogen becomes:
Vn / Vtot = Pn / Ptot
78/100 = Pn / 1
Pn = 0.78 atm
Henry constant for N₂ at 25 °c = 1600 atm/mol.L=6.25.10⁻⁴ mol/L.atm
The solubility :

The mass of sodium sulfite that was used will be 1,890 grams.
<h3>Stoichiometric problems</h3>
First, the equation of the reaction:

The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.
Mole of 960 grams SO2 = 960/64 = 15 moles
Equivalent mole of sodium sulfite that reacted = 15 moles
Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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The correct answer is D. Number of grams of salt per kilometer of water.