Handout Lecture 14

7/27/2019 Handout Lecture 14

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Chapter 17: Properties of Solutions

17.1 Solution Composition

17.2 The Thermodynamics of Solution Formation

17.3 Factors Affecting Solubility

17.4 The Vapor Pressure of Solutions

17.5 Boiling-Point Elevation and Freezing-Point Depression

17.6 Osmotic Pressure

17.7 Colligative Properties of Electrolytic Solutions

17.8 Colloids

Figure 17.4: (a) a gaseous solute inequilibrium with a solution. (b) the piston is

pushed in, which increases the pressure of

the gas and the number of gas molecules per

unit volume. (c) greater gas


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Henrys Law of Gas solubilities in Liquids

P = kHX

P = Partial pressure

of dissolved gas

X = mole fraction

of dissolved gas

kH = Henrys Law

Constant

Henrys Law of Gas SolubilityProblem: The lowest level of oxygen gas dissolved in water that will

support life is ~ 1.3 x 10 - 4 mol/L. At the normal atmospheric pressure of

oxygen is there adaquate oxygen to support life?

Plan: We will use Henrys law and the Henrys law constant for oxygen

in water with the partial pressure of O2 in the air to calculate the amount.

Solution:

Soxygen = kH x PO2 = 1.3 x 10-3 mol x ( 0.21 atm)

liter atm

SOxygen = 2.7 x 10- 4 mol O2 / liter

.

The Henrys law constant for oxygen in water is 1.3 x 10-3 mol

liter atm

and the partial pressure of oxygen gas in the atmosphere is 21%,

or 0.21 atm.

.

This is adaquate to sustain life in water!


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Figure 17.5: The solubilities of several

solids as a function of temperature.

Predicting the Effect of Temperatureon Solubility - I

Problem: From the following information, predict whether the

solubility of each compound increases or decreases with an increase in

temperature.

(a) CsOH Hsoln = -72 kJ/mol

(b) When CsI dissolves in water the water becomes cold

(c) KF(s) K+

(aq) + F-(aq) + 17.7 kJ

Plan: We use the information to write a chemical reaction that includesheat being absorbed (left) or released (right). If heat is on the left, a

temperature shifts to the right, so more solute dissolves. If heat is on

the right, a temperature increase shifts the system to the left, so less

solute dissolves.

Solution:

(a) The negative H indicates that the reaction is exothermic, so when

one mole of Cesium Hydroxide dissolves 72 kJ of heat is released.

H2O


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Predicting the Effect of Temperature

on Solubility - II

(a) continued

CsOH(s) Cs+

(aq) + OH-(aq) + Heat

A higher temperature (more heat) decreases the solubility of CsOH.

H2O

(b) When CsI dissolves, the solution becomes cold, so heat is absorbed.

CsI(s) + Heat Cs+

(aq) + I-(aq)

H2O

A higher temperature increases the solubility of CsI.

(c) When KF dissolves, heat is on the product side, and is given off

so the reaction is exothermic.

KF(s) K+

(aq) + F-(aq) + 17.7 kJ

H2O

A higher temperature decreases the solubility of KF

Figure 17.6:The solubilities

of several gases

in water

as a function of

temperature at a

constant

pressure of1 atm of gas

above the

solution.


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Figure 17.9: The presence of a nonvolatile

solute inhibits the escape of solvent

molecules from the liquid

Figure 17.10: For a solution that obeysRaoults law, a plot of Psoln versus xsolvent

yields a straight line.


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Fig. 13.15

Vapor Pressure Lowering -I

Problem: Calculate the vapor pressure lowering when 175g of sucrose

is dissolved into 350.00 ml of water at 750C. The vapor

pressure of pure water at 750C is 289.1 mm Hg, and its

density is 0.97489 g/ml.

Plan: Calculate the change in pressure from Raoults law using the

vapor pressure of pure water at 750C. We calculate the mole

fraction of sugar in solution using the molecular formula of

sucrose and density of water at 750

C.Solution: molar mass of sucrose ( C12H22O11) = 342.30 g/mol

175g sucrose

342.30g sucrose/mol= 0.51125 mol sucrose

350.00 ml H2O x 0.97489g H2O = 341.21g H2O

ml H2O 341.21 g H2O

18.02g H2O/mol= 18.935 molH2O


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Vapor Pressure Lowering - II

Xsucrose =mole sucrose

moles of water + moles of sucrose

Xsurose = = 0.26290.51125 mole sucrose

18.935 mol H2O + 0.51125 mol sucrose

P = Xsucrose x P0H2O = 0.2629 x 289.1 mm Hg = 7.600 mm Hg

Like Example 17.1 (P 841-2)A solution was prepared by adding 40.0g of glycerol to 125.0g of water

at 25.0oC, a temperature at which pure water has a vapor pressure of

23.76 torr. The observed vapor pressure of the solution was found to be

22.36 torr. Calculate the molar mass of glycerol!

Solution:

Roults Law can be rearranged to give:

XH2O = = = 0.9411 =

mol H2O = = 6.94 mol H2O

0.9411 =

mol gly = = 0.4357 mol

Psoln

P

o

H2O

22.36 torr

23.76 torr

mol H2O

mol gly + mol H2O125.0 g

18.0 g/mol6.94 mol

mol gly + 6.96 mol6.94 mol (6.94 mol)(0.9411)

0.9411

40.0 g

0.4357 mol= 91.81 g/mol (MMglycerol = 92.09 g/mol)

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Handout Lecture 14