If 0.580mol of a nonvolatile nonelectrolyte are dissolved in 3.50mol of water, w

Question

If 0.580mol  of a nonvolatile nonelectrolyte are dissolved in 3.50mol of water, what is the vapor pressure PH2O of the resulting solution? The vapor pressure of pure water is23.8 torr at 25 ?C .

In solutions composed of two liquids (A and B), each liquid contributes to the total vapor pressure above the solution. The total vapor pressure is the sum of the partial pressures of the components:

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A 2.500×10^-2M solution of NaCl in water is at 20.0?C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.3mL . The density of water at 20.0?C is 0.9982 g/mL.

a)Calculate the molality of the salt solution.

b)Calculate the mole fraction of salt in this solution.

c)Calculate the concentration of the salt solution in percent by mass.

d)Calculate the concentration of the salt solution in parts per million.

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Atmospheric pressure is 1.01 bar, but air is only composed of 21.0% (by moles) oxygen. Thus, the partial pressure of oxygen in air is less than 1.01 bar. What is the millimolar concentration of dissolved oxygen for a lake or stream under the partial pressure of oxygen and a temperature of 16?C?

Explanation / Answer

1.

P solution = X solvent Po solvent

P solution = Vapor pressure of resulting solution.

X solvent = Mole fraction of solvent = Moles of solvent / Total moles

Moles of water = 3.50 mol

Moles of nonvolatile electrolyte = 0.580 mol

Total moles = 4.08 mol

Mole fraction = 3.50 / 4.08 = 0.86

Po solvent = 23.8 torr

Plugging the values in P solution = X solvent Po solvent

P solution = 0.86 x 23.8 = 20.47 torr

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a)

Molality = Moles of solute / Kg of solvent

Given, 2.5 x 10^-2 M of 1 L solution.

So, moles of NaCl = 2.5 x 10^-2 moles

Given, volume of water = 999.3 mL

Mass = density x volume = 0.9982 g/ml x 999.3 mL = 997.5 g

Molality = (2.5 x 10^-2 mol) / 997.5 = 2.51 x 10^-5 m

b)

Mole fraction of salt = [Moles of salt / Total moles]

Moles of salt = 2.5 x 10^-2

Moles of water = Mass / Molar mass = 997.5 g / 18 g/mol = 55.4 mol

Total moles = 55.442 mol

Mole fraction of salt = (2.5 x 10^-2 ) / 55.442 = 4.51 x 10^-4

c)

Percent by mass = [(mass of solute / Mass of solution)] x 100

Mass of solute = Moles of solute x molar mass = 2.5 x 10^-2 x 58.5 = 1.46 g

Mass of solution = mass of solute + Mass of solvent = 1.46 g + 997.5 g = 998.96 g

Percent by mass = (1.46 / 998.96) x 100 = 0.146 %

d)

ppm = (g of solute/g of solution) x 10^6

ppm = 1.46 / 998.96) x 10^6 = 1461.52 ppm

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the partial pressure of oxygen is 0.21x 1=0.21 atmosphere
The solubility of O2 at 16 C is 9.858 x 10^-3g/L

1 mole of O2 weighs 32 g
so molarity = 0.009858/32=0.000308=3.08 x 10^-4 mole/L = 3.08 x 10^-7 milli moles /L

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