Question

1. The solubility of nitrogen in water at a particular temperature is 6.24x10-4 M. If the temperature of this solution is increased, what is a possible solubility of nitrogen in water?

For the aqueous solution described in number 1, the external pressure was 1 atm. What mass of nitrogen is dissolved in 575 mL of water if the pressure above the solution is increased to 5.75 atm? Would the mass of nitrogen increase or decrease if the pressure above the solution decreased to 0.575 atm?

Answer 1

1. By increasing Temperature(T) solubility of gas decreases.

2.
Henry's law

C = kP

k = Henry's law constant = ? mol/L*atm

P = partial pressure of N2 = 1 atm

C = molar solubility of N2 in water = 6.24*10^-4 mol/l

k = c/p = 6.24*10^-4/1 = 6.24*10^-4 mol/L*atm

now, if P1 = 5.75 , c1 = ?

        p2 = 1 , c2 = 6.24*10^-4

c1/p1 = c2/p2

(x/5.75) = ((6.24*10^-4)/1)

c1 = x = 3.6*10^-3 M

No of mol of N2 present = 3.6*10^-3*0.575 = 0.00207 mol

mass of N2 present = 0.00207*28 = 0.058 g

if the pressure above the solution decreased to 0.575 atm , mass of nitrogen decreases.