Chemistry H: Form TR8.6A

Name ______________________________

TEST 8 REVIEW

Date _________________ Period _____

Test Review # 8 Assumptions of the Ideal Gas Law. This Ideal Gas model is based on the following assumptions, and can be applied only under conditions of low pressure and high temperature: (1) gas molecules are continuously moving in a random, straight line motion. (2) when gas molecules collide with each other or with the walls of the container there is no energy lost. Therefore, the total energy of the system never changes. (3) the actual volume of the molecules is insignificant when compared to the volume of the contained area (the container). (4) no attraction exists between molecules. Gases deviate from the ideal conditions when conditions of high pressure and low temperature exist. These conditions lead to confinement and intermolecular attractions begin to act. In fact, gas molecules do have a volume of their own, and there are forces of attraction between gas molecules. The factors allow for the existence of gases as either solids or liquids under certain conditions,

Gas Stoichiometry. Since the ideal gas law makes it possible to calculate the number of moles of a gas contained in a given volume of gas, it is possible to do the same type of calculations based on a balanced equation that one does with masses.

Sample Problem 1 How many grams of rust (Fe2O3) form when iron reacts with 25.0 L of oxygen at 25EC and 200. kPa?

Sample Problem 2 How many milliliters of ammonia are formed when 150.mL of hydrogen combines with nitrogen at constant temperature and pressure?

Step 1: Write a balanced equation 4Fe + 3O2 ÿ 2Fe2O3 Step 2: Substitute values into the gas equation to get the number of moles of gas (200. kPa )(1atm)(25.0 L) = 2.02mol PV n= = L⋅atm . kPa )(0.0821 mol RT (1013 ⋅ K )(298 K )

Step 1: Write a balanced equation N2 + 3H2 ÿ 2NH3 Step 2: Set up a proportion and solve 3mol H2 2mol NH3 = x = 100ml NH3 150. ml H2 x

Step 3: Solve the remaining problem by the factor label method. ⎛ 2mol Fe2O3 ⎞ ⎛ 159.7 g Fe2O3 ⎞ ⎟⎟ = 215g Fe2O3 ⎟⎟ ⎜⎜ 2.02molO2 ⎜⎜ ⎝ 3molO2 ⎠ ⎝ 1mol Fe2O3 ⎠

Sample Problem 3 How many liters of oxygen are liberated when 18.4 g of potassium chlorate decompose at STP?

G raham’s Law. The rate at which gases effuse is inversely proportional Step 1: Write a balanced equation. to the square root of the molar mass. 2KClO3 ÿ 2KCl + 3O2 MB Step 2: Solve by the factor label method rate of effusion of A = rate of effusion of B MA ⎛ 1mol KClO3 ⎞ ⎛ 3molO2 ⎞ ⎛ 22.4 LO2 ⎞ ⎟⎟ = 5.04 L ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ 18.4 g KClO3 ⎜⎜ Sample Problem 1 ⎝ 122.6 g KClO3 ⎠ ⎝ 2mol KClO3 ⎠ ⎝ 1molO2 ⎠ What is the molar mass of a gas that effuses at 1.25 times the speed of carbon dioxide? M CO2 rate of effusion of A = rate of effusion of CO2 MA 125 . = MA =

44.0 g mol MA 44.0 g mol 2 . ) (125

so (125 . ) =

= 28.2 g mol

2

44.0 g mol MA

Test Review 8

Chemistry H: Form TR8.11A TEST 8 REVIEW

Page 2

Characteristics of Solutions. A solution is a homogeneous mixture. Like mechanical mixtures, a solution is composed of two or more substances and has variable composition BUT the particles are distributed evenly throughout each other SO the composition is uniform and the solution appears to be one substance. A solution consists of a solute dissolved in a solvent. The solute is the substance that is dissolved by another. The solvent is the substance that dissolves another. The solvent is distinguished from the solute in that it is the continuous phase. For example, when salt, a solid, dissolves in water, a liquid, the solution appears to be a liquid, so water is the solvent. Solubility. Solubility is the ability of a solute to dissolve in water or another solvent. Several factors affect solubility. [1] The nature of solute and solvent: In order for a solvent to dissolve a solute, it must exert forces of attraction on the solute. Polar solvents such as water dissolve polar and ionic solutes because they exert mutual attractions that cause their particles to intermingle. Nonpolar solvents such as benzene do NOT dissolve polar and ionic substances because they exert no forces of attraction that would cause the particles to separate so they can intermingle. As an exaple, oil and water do NOT mix. [2] Temperature (See Table G - Solubility Curves): The solubility of solid solutes generally increases as temperature increases while the solubility of gaseous solutes generally decreases as temperature increases. [3] Pressure: The solubility of solids and liquids are not effected by pressure, but the mass of a dissolved gas in a liquid is directly proportional to the pressure of the gas. Rate of solution. The factors that influence the rate of solution differ for solid and gaseous solutes. Factor

Affect on Solid Solute

Affect on Gaseous Solute

Particle Size

reducing particle size by crushing increases the rate by increasing surface area

Stirring

increases the rate by exposing fresh solvent to solute and increasing kinetic energy

decreases the rate by increasing kinetic energy, thereby reducing solubility

Amount of dissolved solute

as the amount of dissolved solute increases, the rate decreases

as the amount of dissolved solute increases, the rate decreases

Temperature

as the temperature increases, the rate increases

as the temperature increases, the rate decreases

Pressure

none

as the pressure increases, the rate inreases

not applicable

Saturation (see Table G). A saturated solution is a solution that cannot dissolve any more solute at a given temperature. Added solute will NOT dissolve in a saturated solution. An unsaturated solution is a solution that can dissolve more solute at a given temperature. Added solute will dissolve in an unsaturated solution. A supersaturated solution is a solution that holds more solute than it can dissolve at a given temperature. It is produced by dissolving solute at a high temperature and allowing it to cool slowly. Addition of solute to a supersaturated solution causes precipitation of the excess. Concentration. Concentration is the amount of solute compared to mass ( solute) solvent. A concentrated solution has a large amount of solute compared to × 100% A.. percent mass = mass ( solution) the amount of solvent, while a dilute solution has a small amount of solute compared to the amount of solvent. Concentration is calculated as shown volume ( solute) × 100% B.. percent volume = in the box to the right. Molarity (M) is a measure of concentration as well. volume ( solution) The formulas for calculating molarity as well as the other variables related mass( solute) to molarity are shown in the box to the right. Concentration can also be × 1,000,000 ppm C.. ppm = expressed as percent solution and parts per million (ppm). The formulas for mass( solution) these calculations are shownto the right.

Molarity. molarity.

Following are formulas showing how to calculate Mass of solute( g ) Volume of Solvent or Solution(mL) •. M = moles( solute) L( solution) g < M= GFM × L < moles = M × L

•. Concentration =

moles M