### Richard A. Paselk

Chem 107

Fundamentals of Chemistry

Fall 2009

Lecture Notes: 17 November

PREVIOUS

NEXT

Returned Exam 2

# Solution Concentrations-a Review & Some New Stuff.

Solutions: a solution occurs when one chemical is completely dissolved or dispersed in another. We most commonly think of solutions as being liquid, but solid solutions also occur, such as the various metal alloys like steel, brass and bronze.

In a solution the substance present in highest concentration is considered to be the solvent, while components in lesser amounts are considered to be solutes. If you dissolve a sugar cube in water you get a sugar solution, where water is the solvent, and sugar is the solute.

### FYI

Example:

• What is the solvent in 80 proof rum: 80 proof = 40% alcohol in water, so water is the solvent.
• What is the solvent in 151 proof rum: 151 proof = 75.5% alcohol in water, so alcohol is the solvent.

## Concentration Measures

### Concentration Terms

Percent Concentration

Mass percent

ppt = parts/thousand (1mg/L of water); ppb = parts/billion (1 microgram/L of water)

Volume percent

Molarity: The most commonly used concentration term in chemistry = moles of solute dissolved in 1 L of solution.

Two types of situation arise giving two kinds of problems:

Making molar solutions.

### FYI

Example: Make up a 1.00000 L solution of 0.25 M NaCl (note that water is the "default" solvent).

First weigh out o.25 moles of NaCl

= (0.25 mole)(22.99 g + 35.45 g)/mole = 14.61 g

Example: What is the concentration of a solution made by dissolving 10.00 g of KI in enough water to make
1.00000 L?

First need to find the number of moles of KI:

(10.00 g) / ({39.10 g+ 126.9 g}/mole) = 6.135 x 10-2 mole

Thus the concentration will be 6.135 x 10-2 M

Dilution problems (see 22 September).

Molality: = moles of solute dissolved in 1 kg of solvent.

Example: What is the concentration of a solution made by dissolving 10.00 g of KI in 1.00000 kg of water?

First need to find the number of moles of KI:

(10.00 g) / ({39.10 g+ 126.9 g}/mole) = 6.135 x 10-2 mole

Thus the concentration will be 6.135 x 10-2 m

Mole fraction: = moles of solute dissolved in total moles of solution = na /n

Example: What is the mole fraction of a solution of 10.0 moles of glycerol dissolved in 15.0 moles of water?

(10 mol) / (10 mol + 15 mol) = 10/25 = 0.400

## Solubility

• All gases are completely soluble in each other.
• Liquid solutions
• "Like dissolves like."
• Solvation - the process of surrounding a solute with solvent molecules).
• Hydration (solvation with water).
• Saturation - the maximum amount of solute which can be in solution in equilibria with its pure state.
• Supersaturation - dissolving solute in excess of saturation - unstable to the addition of solute (carbonated beverages, honey, etc.) Video
• Temperature effects:
• Gases decrease in solubility with increasing temperature. (example of oxygen solubility)
• Many solids increase in solubility with increasing temperature, but not always true.
• Pressure effects:
• Gases increase in solubility with increasing pressure (example of carbonated beverages)
• Solubility of solids generally uneffected by pressure, since both liquids and solids essentially incompressible.

Colloids: defined by particle size = 1.0 nm< colloid < 100 nm (particles in solution are 0.1 - 1.0 nm in diameter, whereas particles > 100 nm dispersed in a fluid are considered to be in suspension.) Colloids generally do not settle out.

• Biomacromolecules are often colloidal (proteins, DNA & RNA)
• particles in inks are sometime colloidal.

Colligative properties (properties which depend only on the number or concentration, not on the type, of particles). Be able to solve problems for:

• vapor pressure lowering (Raoult's Law): P = XP°, where P° = the vapor pressure of the pure substance and X = its "mole fraction", that is the number of moles of substance divided by the total number of moles of all substances in the solution (moles solute/(moles solute + moles solvent)) In other words the vapor pressure of a substance in solution is proportional to the molecular percentage of that substance in the solution.
• boiling point elevation: Tb = kbm, where m = molality = moles solute/kg solvent, and kb is a constant specific to the solvent.
• freezing point depression:Tf = -kfm, where m = molality = moles solute/kg solvent, and kf is a constant specific to the solvent. (Recall MW lab as example).
• osmotic pressure ():V = nRT; or, dividing both sides by V = MRT, where M = molarity.
• Example: What are the osmotic pressures of 1.00 M sugar and 1 M aluminum chloride solutions at 25°C?

sugar = MRT = (1 mol/L)(0.0821 L*atm/mol*K)(298 K) = 24.5 atm

AlCl3 = MRT = (1 mol/L)(4 mol ion/mol)(0.0821 L*atm/mol*K)(298 K) = 97.9 atm

 C107 Home