Solutions Review Solutions Test 1A, 2A: Thursday 2/28 3B: Friday 3/1 30 Multiple Choice (60) 40 points short answers

Differences in physical properties such as mass, particles size, molecular polarity, boiling point, and solubility permit physical separation of the components of the mixture. Sometimes they ask about separating components of mixtures. Separations exploit differences in physical properties: solubility (filtration), boiling point (distillation)

A solution is a homogenous mixture of a solute dissolved in a solvent. The solubility of a solute in a given amount of solvent is dependent on the temperature, pressure, and the chemical natures of the solute and solvent. The concentration of a solution is expressed in molarity (M), percent by volume, percent by mass, or parts per million (ppm).

The formulas for molarity and ppm are found on Table T. Molarity is a hard calculation at times because it contains moles in the numerator, which is another calculation. Here are some simple guidelines for molarity calculations: - If moles are given or the unknown, just solve for them using the molarity formula. – If grams are given, you must calculate molar mass and then moles using the mole formula, then plug into molarity formula. – If grams of solute is the unknown, you must start with the molarity formula and determine moles; plug the value for moles back into the moles formula. – Both formulas can be combined using complex fractions; the numerator of the molarity equation is the moles equation.

Practice Molarity Questions How many grams of Ca(NO 3 ) 2 are required to produce 1.5 L of a 0.50 M solution?

How many moles of CuCl 2 are required to make 500 ml of a 1.2 M solution?

What is the concentration of a solution containing 10.5 grams of KCl in 250 ml of water?

Freezing Point Depression Boiling Point Elevation The addition of a nonvolatile solute to a solvent causes the boiling point of the solvent to increase and the freezing point of the solvent to decrease. The greater the concentration of particles, the greater the effect.

Unequal distribution of charge is a euphemism for a polar molecule. These molecules have dipoles [partial (+) and (–) parts of the molecule]. For similar molecules, the larger the dipole, the higher the melting point and boiling point are. Water is a very polar molecule. It has strong dipole-dipole forces, and makes hydrogen bonds with itself and polar solutes.

For a given number of moles of a substance, the ionic compounds have the biggest effect on boiling point and freezing point because they dissociate into ions. The more moles of ions you get, the bigger the effect. CH 3 OH: one mole of methanol molecules NaOH:one mole of Na + and 1 mole of OH - ions Ba(OH) 2 : one mole of Ba 2+ ions and 2 moles of OH - ions. The barium hydroxide has the biggest effect for a given molar concentration of solute.

Ion-dipole interactions make most ionic compounds soluble in water. Be clear on which part of the water molecule is attracted to the cation (+) and anion (-).

Information about the solubility of various compounds is given in Table G. Concentrations are given in grams of solute per 100 g of water. Every example given except three are ionic compounds of relatively high solubility. All of these compounds become more soluble as temperature is increased.

Table F: Solubility Rules The solubility rules of Table F cover a wide range of ionic compounds not found on Table G. Even compounds that are “insoluble” have small amounts of ionization in solution. You must be able to determine whether a product is soluble or insoluble using the table from formulas or names.

Table G questions are common; mostly they involve careful reading of the question and working with the graph. A common question asks for the solubility of a solute in less than or more than 100 g of water. Other common questions involve whether a solution is saturated or unsaturated (below the line and on the line, respectively). The three gases (HCl, NH 3, and SO 2 ) have decreased solubility at high temperatures. This is true for all gases. The solubility of gases is also strongly correlated with atmospheric pressure.

Here is a practice version of what you will see on part 2 of the test. NameFormulaS or IJustification Iron (III) nitrate CaCrO 4 PbBr 2 Ammonium Carbonate