Tugas Kimia Rutin Part -2

6. The solubility of nitrogen in water is 8.21 x 10^-4 mol/L at 0°C when the N₂ pressure above water is 0.790 atm. Calculate the Henry’s law constant for N₂ in units of L.atm/mol for Henry’s law in the form P = kC, where C is the gas concentration in mol/L. Calculate the solubility of N₂ in water when the partial pressure of nitrogen above water is 1.10 atm at 0°C.

Answer :

C = kP

Rearrange to solve k

k = C/P

k = 8.21 x 10^-4 mol/L / 0,790 atm = 1.04 x 10^-3 mol/atm.L

Then, solve for C

C = 1.04 x 10^-3 molL^-1atm^-1 x 1.10 atm = 1.14 x 10^-3 mol/L

7. Calculate the solubility of O₂ in water at a partial pressure of O₂ of 120 torr at 25°C. the Henry's law constant for O₂ at 25°C is 1.3 x 10^-3 mol/L atm. How do you expect the solubility to change if the temperature were decreased?

Answer :

P = 120 torr/760 = 0,158 atm

Cgas = kP

= (1.3 x 10^-3 mol/L atm) (0.158 atm)

= 2,054 x 10^-4 mol/L.

Decreasing the temperature will increase the solubility of O₂ gas in water.

When the temperature increases, the solubility of O₂ gas in water will decrease because the increase in temperature will increase the kinetic energy of gas particles and increase its motion that will break intermolecular bonds and escape from solution.

8. Glycerin, C₃H₈O₃, is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g of glycerin to 342 mL of H₂O at 39.8°C? The vapor pressure of pure water at 39.8°C is 54.74 torr and its density is 0.992 g/cm³.

Answer :

moles glycerin = 164 g / 92.09 g/mol =1.78

assuming density of water = 1 g/mL

mass water = 342 mL x 1 g/mL = 342g

moles water = 342 g/ 18.02 g/mol = 19.0

mole fraction water = 19.0 / (19.0 + 1.78) = 0.914

P = P° X

P = 54.74 x 0.914 = 50.05 torr

9. The vapor pressure of a solution containing 53.6 g glycerin (C₃H₈O₃) in 133.7 g ethanol (C₂H₅OH) is 113 torr at 40°C. Calculate the vapor pressure of pure ethanol at 40°C assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.

Answer:

moles glycerin = 53.6 g /92 g/mol = 0.582

moles ethanol = 133.7 g /46 g/mol = 2.91

total moles = 3.492

Moles fraction ethanol (X) = 2.91 / 3.492 = 0.833

P = P°.X

113 = P° 0.833

P° = 135.6 torr

10. The normal boiling point of methanol is 64.7 °C. A solution containing a nonvolatile solute dissolved in methanol has a vapor pressure of 450.7 torr at 64.7°C.

What is the mole fraction of methanol in this solution?

Answer :

P solvent = P°. X solvent

450,7 = 760 . X solvent

X solvent = 450,7/760 = 0,593

X solute = 1 - 0,593 = 0,407

Source : slader.com
For number 1-5 Read HERE
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5NChemistry