II

III

I II. Solution Concentration

(p. 480 – 488)

II. Solution Concentration

(p. 480 – 488)

Ch. 14 – Mixtures & SolutionsCh. 14 – Mixtures & Solutions

A. ConcentrationA. Concentration

The amount of solute in a solution

Describing Concentration

• % by mass - medicated creams

• % by volume - rubbing alcohol

• ppm, ppb - water contaminants

• molarity - used by chemists

• molality - used by chemists

B. Percent by MassB. Percent by Mass

Percent Composition by Mass is the mass of the solute divided by the mass of the solution (mass of the solute plus mass of the solvent), multiplied by 100.

Mass of SoluteMass of Solution

x100

B. Percent by MassB. Percent by Mass

How many moles of solute are contained in 343 grams of a 23% aqueous solution of MgCr2O7?

343g of solution 23%

100%

1 mol

240.3 g MgCr2O7

=0.329 mol of MgCr2O7

C. Percent by VolumeC. Percent by Volume

Percent Composition by Volume is the volume of the solute divided by the volume of the solution (volume of the solute plus volume of the solvent), multiplied by 100.

Volume of SoluteVolume of Solution

x100

B. Percent by VolumeB. Percent by Volume

Determine the percent by volume of toluene (C6H5CH3) in a solution made by mixing 40.0 mL of toluene with 75.0 mL of benzene (C6H6).

40.0 mL toluene + 75.0 mL benzene= 115 mL total

solution(40.0 mL toluene / 115 mL solution) 100 = 34.8%

toluene

C. MolarityC. Molarity

Concentration of a solution

solution of liters

solute of moles(M)Molarity

total combined volume

substance being dissolved

C. MolarityC. Molarity

2M HCl

L

molM

nsol' L 1

HCl mol 2HCl 2M

What does this mean?

C. Molarity CalculationsC. Molarity Calculations

How many grams of NaCl are required to make 0.500L of 0.25M NaCl?

0.500 L sol’n 0.25 mol NaCl

1 L sol’n

= 7.3 g NaCl

=.125 mol NaCl

58.44 g NaCl

1 mol NaCl

.125 mol NaCl

C. Molarity CalculationsC. Molarity Calculations

Find the molarity of a 250 mL solution containing 10.0 g of NaF.

10.0 g NaF 1 mol NaF

41.99 g NaF

L

molM .283 NaF

.25 L sol’n= 0.95 M

NaF

=.283 mol NaF

D. MolalityD. Molality

solvent ofkg

solute of moles(m)molality

mass of solvent only

1 kg water = 1 L waterkg 1

mol0.25 0.25m

D. MolalityD. Molality

Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.

75 g MgCl2 1 mol MgCl2

95.21 g MgCl2

= 3.2m MgCl2

0.25 kg water

kg

molm

D. MolalityD. Molality

How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water?

0.500 kg water 1.54 mol NaCl

1 kg water

= 45.0 g NaCl

58.44 g NaCl

1 mol NaCl

kg 1

mol1.54 1.54m

E. Mole FractionE. Mole Fraction

The number of moles of a component of a solution divided by the total number of moles of all components.

Moles A

Total Moles= Mole Fraction, Χ

2211 VMVM

C. DilutionC. Dilution

Preparation of a desired solution by adding water to a concentrate.

Moles of solute remain the same.

C. DilutionC. Dilution

What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?

GIVEN:

M1 = 15.8M

V1 = ?

M2 = 6.0M

V2 = 250 mL

WORK:

M1 V1 = M2 V2

(15.8M) V1 = (6.0M)(250mL)

V1 = 95 mL of 15.8M HNO3

E. Preparing SolutionsE. Preparing Solutions

500 mL of 1.54M NaCl

500 mLwater

45.0 gNaCl

• mass 45.0 g of NaCl• add water until total

volume is 500 mL• mass 45.0 g of NaCl• add 0.500 kg of water

500 mLmark

500 mLvolumetric

flask

1.54m NaCl in 0.500 kg of water

E. Preparing SolutionsE. Preparing Solutions

250 mL of 6.0M HNO3 by dilution

• measure 95 mL of 15.8M HNO3

95 mL of15.8M HNO3

water for

safety

250 mL mark

• combine with water until total volume is 250 mL

• Safety: “Do as you oughtta, add the acid to the watta!” or AA – add acid!