Chap13 - Solutions and Colligative Properties A solution is a solute (A) dissolved into a solvent...

Chap13 - Solutions and Chap13 - Solutions and Colligative PropertiesColligative Properties

A solution is a solute (A) dissolved A solution is a solute (A) dissolved into a solvent (B).into a solvent (B).

Solvent B

Solute A

A. ConcentrationA. Concentration

1. Mass Percent1. Mass Percent

= = mass component mass component x 100 x 100 total mass solution total mass solution

= = grams A grams A x 100 x 100

grams A + grams Bgrams A + grams B

2. Parts per million (ppm)2. Parts per million (ppm)

= = mass componentmass component x 10x 1066

total mass solutiontotal mass solution

ppt = x 10ppt = x 1033 ( ) ( )

ppb = x 10ppb = x 109 9 ( )( )

Parts per thousand

Parts per billion

3. Mole fraction (X)3. Mole fraction (X)

= = mole componentmole component

total molestotal moles

XXAA = mole A = mole A

mole A + mole Bmole A + mole B

4. Molarity (M)4. Molarity (M)

= = moles of solutemoles of solute

Liters of solutionLiters of solution

M = mole AM = mole A

L solutionL solution

3.16 g MgBr2 1 mol = 0.0172 mol MgBr2

184.1 g MgBr2

*Remember*1L = 1000mL

0.859 L

Look at Problem #1

= 0.0200 M

5. Molality (m)5. Molality (m)

= = moles solutemoles solute

kg solventkg solvent

m = m = mole Amole A

kg Bkg B *Remember*1g = 1mL for H2O 1000g = 1 kg

4.8 g NaCL 1 mol = 0.0821 mol NaCl

58.5 g NaCl 0.5 kg

Look at Problem #4

= 0.164 m

Dilution of SolutionDilution of Solution

MM11VV1 1 = M= M22VV22

Answer to Worksheet

3a – 50 mL diluted3b – 1.28 mL diluted

Conversion between UnitsConversion between Units

For HFor H22O only, Molarity = molality.O only, Molarity = molality.

Why? Because the density of HWhy? Because the density of H22O is equal O is equal

to 1.00 g/mL.to 1.00 g/mL.

Therefore, 1000mL = 1000gTherefore, 1000mL = 1000g 1 L = 1 kg1 L = 1 kg

Conversion between UnitsConversion between Units

For any other solution other than an For any other solution other than an aqueous solution - aqueous solution - YOU MUST YOU MUST

USE THE USE THE DENSITYDENSITY!!!!!!!!!!

Use the density to convert mass to Use the density to convert mass to volume.volume.

Conversion between UnitsConversion between Units

mass solvent

mass solute

mass solution

volume solution

moles solutemolarmass

M = mole/L

m = mole/kg

density

+ add

Conversion between UnitsConversion between Units

mass solvent

mass solute

mass solution

volume solution

moles solutemolarmass

M = mole/L

m = mole/kg

density

+ add

1000 g = 1 kg

1000 mL = 1 L

Conversion between UnitsConversion between Units

molarmass

density

+ add

1000 g = 1 kg

1000 mL = 1 L

Your homework/classwork is Your homework/classwork is worksheet –worksheet –

concentration conversionsconcentration conversions

Convert mass % to …..Convert mass % to …..

5% HC5% HC22HH33OO22 5 g x 1mol/60g = 5 g x 1mol/60g =

0.0833 mol0.0833 mol

95% H95% H22O O 95 g x 1mol/18g = 95 g x 1mol/18g =

5.28 mol5.28 mol

Mole fractionMole fraction

X = X = 0.0833mol0.0833mol / ( / (0.0833 mol0.0833 mol + + 5.28 mol5.28 mol))

m = m = 0.0833 mol0.0833 mol / / 0.095 kg0.095 kg

molalitymolality

convert 95 g

Convert M to mConvert M to m

1.13 1.13 mol mol to to mol mol L L solutionsolution kg kg solventsolvent

1000 mL x 1.05 g/ml = 1050 g solution1000 mL x 1.05 g/ml = 1050 g solution

1.13 mole KOH x 56.1 g/mol = 63.4 g solute1.13 mole KOH x 56.1 g/mol = 63.4 g solute

density

1050 g solution 1050 g solution

– – 63.4 g KOH63.4 g KOH

986.6 g solvent986.6 g solvent

m = m = 1.13 mol KOH 1.13 mol KOH = 1.15= 1.15 mol mol

0.9866 kg kg0.9866 kg kg

What about mass percent?What about mass percent?

1050 g solution

63.4 g KOH

100*solution g

solute g % Mass

100*solution g 1050

KOH g 63.4 % Mass

Dimensional AnalysisDimensional Analysis

What is the molarity of concentrated HCl?What is the molarity of concentrated HCl? 39.0% HCl by mass and 1.13 g/mL density39.0% HCl by mass and 1.13 g/mL density

ML

mL

mL

g

g

Mole

sog

HClg1.12

1

1000*

1

13.1*

5.36

1*

ln100

0.39

Solution CalculationsSolution Calculations

What is the molarity What is the molarity of a 1.11 ppm solution of a 1.11 ppm solution of Znof Zn2+2+ ions? ions?

MXg

Znmole

mg

g

liter

Znmg 52

1070.139.65

1*

1000

1*

1

11.1

Solid CalculationsSolid Calculations

Chemical analysis showed 1.23 mg Fe in a Chemical analysis showed 1.23 mg Fe in a

15.67 g sample of soil.15.67 g sample of soil. What is the Fe concentration in ppm?What is the Fe concentration in ppm?

ppmmg

g

sampleg

Femg5.7810*

1000

1*

67.15

23.1 6

Unusual concentration unitsUnusual concentration units

How many nano moles of Cu are present in 12.3 How many nano moles of Cu are present in 12.3 µL of 25 ppm CuSOµL of 25 ppm CuSO44??

molenmolem

molen

mg

molem

L

mg

L

LL 93.1

1

10*

65.159

1*

25*

10

1*3.12

6

6

B. Colligative PropertiesB. Colligative Properties

1. Boiling Point Elevation1. Boiling Point Elevation

ΔTΔTbb = k = kb b • m • • m • ii

for an aqueous solutionfor an aqueous solution

TTbb = 100 = 100ooCC + (0.52 + (0.52 ooC/m) •(m)C/m) •(m)

* note that as molality increases * note that as molality increases

ΔTΔTb b increases as wellincreases as well

Normal B.P. Kbfor water

B. Colligative PropertiesB. Colligative Properties

2. Freezing Point Depression2. Freezing Point Depression

ΔTΔTff = k = kf f • m • • m • ii

for an aqueous solutionfor an aqueous solution

TTff = 0 = 0ooCC - (1.86 - (1.86 ooC/m) •(m)C/m) •(m)

* note that as molality increases * note that as molality increases

ΔTΔTf f increases as wellincreases as well

Normal F.P. Kf for water

Ex. Non-electrolyte (Ex. Non-electrolyte (ii=1)=1)

Antifreeze is made at 25% CAntifreeze is made at 25% C22HH66OO22 by mass. by mass.

What is the T What is the Tb b and the Tand the Tff??

Make your life easy and assume 1000g.Make your life easy and assume 1000g. Why? Because molality is based upon kg of Why? Because molality is based upon kg of

solventsolvent Mass percentMass percent 250 g C 250 g C22HH66OO22

750 g H750 g H22OO

Boiling and Freezing PointBoiling and Freezing Point

mkg

mole

g

molegm 37.5

750.0

03.4

1.62

1250

CmmCCTb 8.102)37.5()/52.0(100

molalitymolality

CmmCCTf 10)37.5()/86.1(0

C2H6O2 H2O

Ex. Molecular Weight of UnknownEx. Molecular Weight of Unknown

What is the MM of a sample if 250grams of What is the MM of a sample if 250grams of the sample is placed into 1000grams of the sample is placed into 1000grams of water and the temperature rose by 3.5water and the temperature rose by 3.5°C?°C?

)(?)/52.0(1005.103 mmCCC

mkg

mole?73.6

Assuming 1000g (1kg), the molality Assuming 1000g (1kg), the molality becomes…..becomes…..

molekgkg

mole73.6173.6

moleMM

g73.6

?

250

molegMM /37

Ex. Electrolyte (Ex. Electrolyte (ii = ?) = ?) IMPORTANT – the colligative properties IMPORTANT – the colligative properties

of freezing point and boiling point are of freezing point and boiling point are proportional to the number of particles proportional to the number of particles present in the solution.present in the solution.

van Hoft factor, van Hoft factor, ii NaCl NaCl ii = 2 moles = 2 moles Ie. 1m = 2m Ie. 1m = 2m CaClCaCl22 i = i = 3 moles3 moles Ie. 1m = 3m Ie. 1m = 3m

AlAl22(SO(SO44))33 i = i = 5 moles5 moles Ie. 1m = 5m Ie. 1m = 5mincreasingcolligative

effect