1. The solubility of O 2 in water is approximately 0.0380 g L -1 of water when t

Question

1. The solubility of O2 in water is approximately 0.0380 g L-1 of water when the temperature is 25.0°C and the partial pressure of gaseous oxygen is 660 torr. The oxygen gas above the water is replaced by air at the same temperature and pressure, in which the mole fraction of oxygen is 0.210. What will the solubility of oxygen in water be under these new conditions?

2. A solution, which was made by dissolving 62.07 g of a nonelectrolyte in 600 g of water, exhibits a freezing point of 1.86°C. What is the molecular weight of this nonelectrolyte compound? For water, Kf is 1.86°C m-1 and Kb is 0.51°C m-1

3. What is the freezing point of a solution that contains 20.0 g of Sodium Chloride in 100.0 g of H2O? Kf for water is 1.86C/m.

Explanation / Answer

1. solubility of O2 = 0.038 g/L

under the same conditions,

solubility in the presence of air   = 0.038*0.21 = 0.00798 g/L

2.

DTf = i*kf*m

1.86 = 1*1.86*(62.07/x)*(1000/600)

x = molarmass of solute = 103.45 g/mol

3.

DTf = i*kf*m

(0-x) = 2*1.86*(20/58.5)*(1000/100)

x = freezing point of a solution = -12.717 C

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