Question

Problem 3 [10 points] (a) Using the following table of Henry's law constants (kB(atm)-PB/XB) calculate the solubility (in M) of each gas in water at 25°C if PO2-0.2 atm , PNzー0.75 atm, and Pco2-0.05 atm Gas Temp. (25°c) N2 O2 CO2 85 × 103 43 × 103 1.61 x 103 b) What will the vapor pressure of water be in this solution if Raoult's law holds? The vapor pressure of pure water at this temperature is 23.756 torr

Answer 1

a)

for O2:

[O2] = 1/H*P-O2 = 1/(43*10^3) * 0.2 = 0.000004651 M

for N2

[N2] = 1/H*P-N2 = 1/(85*10^3) * 0.75= 0.00000882352 M

for N2

[CO2] = 1/H*P-CO2 = 1/(1.61*10^3) * 0.05 = 0.00003105 M

b)

find vpaor pressure:

Ptotal =x1*P1 + x2*P2 + x3*P4

Vapor pressure of water = 23.756 torr then

Pvapor = 23.756 torr