AP Chemistry : Solutions and States of Matter
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Example Questions
Example Question #2 : Solids And Liquids
Boiling points at standard temperature and pressure:
Methanol-
Toluene-
Water-
Methylene chloride-
Dimethylsulfoxide-
Of the five compounds given, which will have the highest vapor pressure at standard temperature and pressure?
Toluene
Methanol
Dimethylsulfoxide
Methylene chloride
Water
Methylene chloride
A liquid will boil when its vapor pressure increases to match the pressure of its environment. By this logic, a liquid with a higher vapor pressure will be closer to boiling. Liquids with low boiling points thus have greater vapor pressures.
Of the given compounds, methylene chloride has the lowest boiling point, meaning it will also have the highest vapor pressure.
Example Question #1 : Properties Of Liquids
Oil and vinegar is a very popular salad dressing. It also is commonly used for dipping bread. The main component of the oil phase is olive oil, while the main part of the vinegar portion is aqueous acetic acid.
When oil and vinegar salad dressing is allowed to stand at room temperature, two distinct phases are observed. The main explanation for this phenomenon is the difference in _________ of the two phases.
solubility
polarity
buoyancy
acidity
density
polarity
Olive oil, like most oils, is non-polar, while aqueous acetic acid is very polar. These two phases do not mix because of their different solvent polarities.
Example Question #21 : Gases
What is the final pressure of a gas initially has a pressure of 10 atm at 50 L if the volume s now 25 L?
20 atm
10 atm
5 atm
25 atm
50 atm
20 atm
Use P1V1 = P2V2
P1 = 10atm; V1 = 50L
P2 = X; V2 = 25L
(10atm)(50L) = (x)(25L)
500 = 25x
x = 20
Example Question #1 : Ideal Gas Law
A sample of ideal gas is heated in a 2L vessel at a temperature of 320 Kelvin. The pressure in the vessel is 2.5atm. What is the new pressure in the vessel if the volume is halved and the temperature is reduced to 250 Kelvin?
Since the gas is ideal, we can use a variation of the ideal gas law in order to find the unknown final pressure.
Since we know that the number of moles is constant between both vessels (and R is a constant as well), we can simply compare the three factors being manipulated between the two vessels: pressure, volume, and temperature. Using a combination of Boyle's law and Charles's law, we can compare the two vessels to one another using the following equation.
Use the given values to solve for the final pressure.
Example Question #1 : Ideal Gas Law
An ideal gas takes up 60L at 2 atm. If the gas is compressed to 30L, what will the new pressure be?
Ideal gas law (modified)
P1V1 = P2V2
P1 = 2 atm; V1 = 60L; P2 = ?; V2 = 30L
(2)(60) = (X)(30)
P2 = 4 atm
Example Question #22 : Gases
Ammonia is created according to the balanced equation below.
The reaction is allowed to take place in a rigid container. Eight moles of hydrogen gas are mixed with two moles of nitrogen gas. The initial pressure exerted on the container is 5atm.
Assuming the reaction runs to completion, what will the pressure exerted on the vessel be after the reaction takes place?
Since the total pressure is dependent on the number of moles in the container, we can use the ratio of moles before and after the reaction to determine the final pressure in the container. There are initially ten moles of gas in the container, eight moles of hydrogen and two moles of nitrogen.
The next step is to determine how many moles of ammonia are created in the reaction, and if there is any excess reactant left over after the reaction. Since there is a 1:3 ratio for hydrogen gas to nitrogen gas, only six of the moles of hydrogen gas will be used in order to react will all two moles of the nitrogen gas.
This leaves two moles of excess hydrogen gas. Using stoichiometry and the molar ratios, we determine that four moles of ammonia are created in the reaction that comsumes two moles of nitrogen.
Four moles of ammonia, plus the two remaining moles of hydrogen gas, results in six moles of total gas after the reaction has run to completion.
Six moles is 60% of the inital moles in the container, so the final pressure will be 60% of the initial pressure. We can solve using the ideal gas law.
Example Question #3 : Ideal Gas Law
How many moles of Oxygen gas are present at a volume of 10 L at 1 atm and 25o C? (MW Oxygen gas = 32 g/mol)
0.041 mol
14 mol
0.41 mol
2.5 mol
4 mol
0.41 mol
use PV = nRT
n = PV/ RT
P = 1 atm; V = 10 L; R = 0.0821 Latm/molK; T = 298 K (MUST switch temperature to K)
n = moles of gas
n = (1atm)(10L)/(0.0821Latm/molK)(298K)
n= 0.41 mol
Example Question #4 : Ideal Gas Law
How many moles will a gas that is behaving ideally be if it takes up 2L at 4atm at 37o C?
0.314 mol
0.6 mol
2.63 mol
3.18 mol
0.379 mol
0.314 mol
Use PV = nRT
n = PV/RT
= (2L)(4atm) / (0.0821 Latm/molK)(310 K) <-- must change T into K
= 8/25.451
= 0.314 mol
Example Question #1 : Ideal Gas Law
If the pressure of a sample of one mole of an ideal gas is increased from 2atm to 3atm at a constant volume, and the initial temperature was 20˚C, what is the final temperature of the sample?
195.3K
439.5K
303K
30K
439.5K
Because the mass and volume of the sample of the ideal gas are kept constant, a change in pressure causes only a direct change in the temperature. This can be derived from the following ideal gas equation:
Example Question #1 : Ideal Gas Law
How many moles of carbon dioxide occupy a space of 16 liters at a pressure of 760torr and a temperature of 250K?
We can calculate the number of moles using the ideal gas law:
Units always have to be in SI units: atm, liters, kelvin, etc. For this question, we have SI units for everything except for the pressure, which is given in torr, rather than atm.
Now that we have the proper units, we can use our given values for pressure, temperature, and volume to find the moles of gas present.
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