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Example Questions
Example Question #11 : Solutions
A volatile solute with a vapor pressure of 23mmHg is added to a solvent with a vapor pressure of 85mmHg. What is the molar fraction of the solvent if the solution has a vapor pressure of 68mmHg?
Since the solute is volatile, it will contribute to the total vapor pressure of the solution. As a result, we must incorporate it when solving for the total vapor pressure.
We can find the vapor pressures using Raoult's law.
Since we are solving for the molar fraction of the solvent, we will designate its molar fraction as . The sum of the molar fractions of each component must be equal to 1; thus, the molar fraction of the solute must be . Using these variables and the information given, we can solve for the molar fraction of the solvent.
Example Question #371 : Ap Chemistry
What is the freezing point of a solution with of sodium chloride in of water?
The equation for freezing point depression is , where is the change in temperature, is a constant related to the solvent, is molality, and is the van't Hoff factor, which is the number of ion particles from each dissolved molecule. We simply plug these numbers into the equation to find the new freezing point.
We know our constant is . Molality is moles of solute per kilogram of solution, and we know that the density of water is one kilogram per liter and the molecular weight of sodium chloride is .
The van't Hoff factor is . Sodium chloride creates only two ions when dissolved, one and one .
Using these values, we can solve for the freezing point depression.
The freezing point will be decreased by . The normal freezing point is , making the new freezing point .
Example Question #13 : Colligative Properties
What is the melting point of a aqueous solution that contains of ?
Since a salt has been added to the pure water, we can find the new melting point of the solution by using the freezing point depression equation:
The change in temperature is equal to the freezing point constant for the solvent multiplied by the molality of the solution multiplied by the van't Hoff factor. The van't Hoff factor is the number of ions that a salt will dissociate into when in solution.
For this particular salt, the van't Hoff factor is three. The molality will be equal to the moles of solute over the mass of the solvent.
Using these terms together in the original equation, we can find the freezing point depression.
This is the change in temperature from the regular freezing point. Since the freezing point for pure water is , the new melting point is .
Example Question #12 : Solutions
A 100mL solution is composed of 25% ethanol by volume and water. What is the mass of the solution?
First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.
Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.
Example Question #1 : Titrations
What is the concentration if 40 g NaOH is dissolved in 1000 g of water
first of all, M = molar; m = molal- M = mol solute/ L of solution; m = mol solute/ kg solvent
you have 40 g NaOH * 1 mol/40 g = 1 mol
1000 g of water is equivalent to 1 L
1 mol/L = 1M
Example Question #1 : Concentration And Units
80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?
0.0016 M
0.04 M
0.625 M
1.6 M
0.00004 M
0.04 M
Molarity = mol solute / L soution
mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol
L solution = 50000 mL water * 1 L/1000 mL = 50 L
2 mol / 50 L =
Example Question #1 : Concentration And Units
How many liters of 10M HCl are needed to make 4L of 0.4M solution?
A simple calculation can be done to perform any solution dilution problem. We know our equation .
We can rewrite this as .
Using this formula, we take the old solution and set it equal to the new solution.
We need 0.16 liters of our 10 molar solution.
Example Question #13 : Solutions
Which solution has the highest molarity?
This question requires us to calculate molarity for each answer choice. It is important to add everything correctly and be careful with more complex compounds.
Molarity is simply moles of solute over liters of solution. The correct answer, after trying each, is the answer with lead (II) nitrate, as it gets us a molarity of 2.
Example Question #3 : Concentration And Units
Which of the following is equivalent to molarity?
Kilograms of solute per mole of solution
Moles of solute per liter of solvent
Moles of solute per liter of solution
Moles of solute per kilogram of solvent
Moles of solute per liter of solution
Molarity, molality, and normality are the three principle ways to measure concentration. Molarity is a measure of moles of solute per liter of solution. Molality is a measure of moles of solute per kilogram of solvent. Normality expressly relates to acids and bases, and is the measure of moles of solute divided by the number of hydrogen equivalents per mole, all divided by liters of solution. Normality is also referred to as "equivalents (of acid) per liter."
Example Question #6 : Concentration And Units
Given that 50 mL of 3M HI was present, what volume would be needed to change the concentration to 0.75M?
150mL
50mL
200mL
180mL
150mL
Use the equation M1V1 = M2V2, and plug in the corresponding values to solve for V2:
V2 = (3)(50)/(0.75) = 200mL
We started out 50mL of the solution, so you have to add 150mL to get a final volume of 200mL.
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