The solubility of N2 in water at a particular temperature and at a N2 pressure of 1 atm is 6.8 * 10-4 mol L-1. Calculate the concentration of dissolved N2 in water under normal atmospheric conditions where the partial pressure of N2 is 0.78 atm.

The correct answer is 5.30 * 10^-4 mol per L.

Explanation:

Based on Henry's law, in a solution solubility of the gas is directly proportional to the pressure, that is, C is directly proportional to P. Here P is the pressure and C is the concentration of the dissolved gases.

Therefore, it can be written as,

C2/C1 = P2/P1

Here, C1 is 6.8 * 10^-4 mol/L, P1 is 1 atm and P2 is 0.78 atm, then the value of C2 obtained by putting the values in the equation,

C2 / (6.8*10^-4) = 0.78/1

C2 = 0.78 * 6.8*10^-4

C2 = 5.30 * 10^-4 mol per L.

Hence, the concentration of dissolved nitrogen at 0.78 atm is 5.30*10^-4 mol/L.