College Chemistry : Liquids and Solutions

Study concepts, example questions & explanations for College Chemistry

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Example Questions

Example Question #1 : Molarity, Molality, Normality

Which of the following choices is characteristic of molality?

Possible Answers:

Useful in experiments without significant temperature changes

Useful in experiments with significant temperature changes

Equivalents per liter

Moles of solute per liter of solution

Correct answer:

Useful in experiments with significant temperature changes

Explanation:

Molarity, molality, and normality are all units of concentration in chemistry. Molarity () is defined as the number of moles of solute per liter of solution. Molality () is defined as the number of moles of solute per kilogram of solvent. Normality () is defined as the number of equivalents per liter of solution. Molality, as compared to molarity, is also more convenient to use in experiments with significant temperature changes. This is because the volume of a solution increases with temperature, and heating causes molarity to decrease; however, since molality is based on masses rather than volumes, molality remains unchanged.

Example Question #2 : Molarity, Molality, Normality

What is the molality of a solution made by adding  of  to  of water?

Possible Answers:

Correct answer:

Explanation:

Molality () is defined as moles of solute per kg of solvent.

 is the solute (it is what is being dissolved) and water is the solvent (what is doing the dissolving).

 Let’s start converting  of  to moles by dividing by its molecular weight using scientific notion through the entire molality calculations as necessary to simplify calculations.

Now we must convert the  of water to  then divide   by it to get the molality:

Molality of solution: 

=

Example Question #1 : Molarity, Molality, Normality

What is the molarity of a solution containing  of solution containing  of ?

Possible Answers:

Correct answer:

Explanation:

Molarity  moles of solute per liter of solution.

We are given that there are  of  and  of solution.

First lets convert the  of solution to liters. It is easiest to use scientific notion in all calculations for easy simplification:

=

Now we must convert  of  to moles to get moles of solute per liter of solution.

This is done by dividing  by its molecular weight which is . Once again we can use scientific notion to simplify calculations:

  

Now we can divide   by  solution to get the molarity.

Example Question #4 : Molarity, Molality, Normality

How many  of water are needed to dilute    to 

Possible Answers:

Correct answer:

Explanation:

In order to solve for the volume of water needed we must use this equation:

solve for  since we are looking for the final volume of water needed to dilute the existing solution:

Example Question #11 : Liquids And Solutions

Which of the following solutions would be expected to have the highest osmotic pressure?

Possible Answers:

Correct answer:

Explanation:

In this question, we're asked to identify an answer choice that would be expected to give us a solution with the greatest osmotic pressure. Remember that osmotic pressure is proportional to the total number of dissolved solute particles in solution, regardless of the identity of those solute particles.

When looking at the answer choices, we need to keep in mind two things. First, we need to recognize the numerical value given for the concentration of the compound given. Secondly, we need to identify if the compound shown is capable of dissociating in solution to give rise to even more solute particles. This is important, as it would affect the osmotic pressure.

 would be expected to have the largest osmotic pressure because, in total, this would be a solution after dissociation occurs.

Example Question #1 : Colligative Properties

Which of the following is not a colligative property?

Possible Answers:

Electronegativity

Vapor pressure depression

Osmotic pressure

Freezing point depression

Correct answer:

Electronegativity

Explanation:

Colligative properties are properties of solutions which depend on the number of dissolved particles in solution. The four main colligative properties are:

1) Freezing point depression: The presence of a solute lowers the freezing point of a solution as compared to that of the pure solvent.

2) Boiling point elevation: The presence of a solute increases the boiling point of a solution as compared to that of the pure solvent.

3) Vapor pressure depression: The vapor pressure of a pure solvent is greater than that of a solution containing a non-volatile liquid. The lowering of vapor pressure leads to boiling point elevation.

4) Osmotic pressure: The osmotic pressure of a solution is the pressure difference between the solution and pure solvent when the two are in equilibrium across a semipermeable membrane. Because it depends on the concentration of solute particles in solution, it is a colligative property.

Electronegativity is not a property of solutions reliant on the number of dissolved particles, but a property of atoms themselves.

Example Question #1 : Colligative Properties

Which of the following aqueous solutions would be expected to have the greatest increase in boiling point?

Possible Answers:

Correct answer:

Explanation:

This question is asking us to identify a solution that increases the boiling point of water by the greatest amount.

To answer this, we need to understand the concept of colligative properties. When a solute dissolves in a solvent such as water, various physical properties are affected. The four colligative properties that change as a result of the addition of solute are freezing point, boiling point, vapor pressure, and osmotic pressure.

With regards to boiling point, as more solute is added to a solution, the boiling point increases. This is due to the fact that addition of solute makes it more difficult for the solute molecules to gain enough kinetic energy at the solution's surface to escape as a gas.

Furthermore, the identity of the solute does not matter. Thus, we need to look only at the number of dissolved solute particles rather than their identity. A compound such as sucrose will not dissociate in solution, which means that the osmotic pressure of the solution is the same as the concentration of sucrose.

Compounds that can dissociate into two or more particles will increase the osmolarity of the solution further. In this case,  will double the stated osmolarity. , on the other hand, will dissociate completely because it is a strong acid, however the protons will not contribute to the osmolarity.

 is able to dissociate into three equivalents of particles in solution. Thus, its initial concentration will be tripled, which gives it the highest osmolarity of any of the choices shown and will thus increase the boiling point by the greatest amount.

Example Question #11 : Liquids And Solutions

What are the products of the following reaction? 

Possible Answers:

Correct answer:

Explanation:

The correct answer is 

First of all, we know this is an acid-base reaction.   is the reactant base and  is the reactant acid.  We know that these acid-base reactions create a water and a salt.  Also, we know that sulfates are soluble, except for those of calcium, strontium, and barium.  As a result,  is a precipitate in the reaction and water is produced.  

Example Question #12 : Liquids And Solutions

Which of the following are soluble? 

Possible Answers:

Phosphates

Carbonates

Sulfides 

Chlorides

Hydroxides

Correct answer:

Chlorides

Explanation:

Most salts containing halogens are soluble.  Carbonates, phosphates, sulfides, oxides, and hydroxides are insoluble except for cations containing alkaline earth metals and hydroxides of calcium, strontium, and barium, which are slightly soluble.  

Example Question #11 : Liquids And Solutions

Calculate the molar solubility of calcium hydroxide. Calcium hydroxide has a  value of .

Possible Answers:

Correct answer:

Explanation:

Start by writing the equation for the dissolution of calcium hydroxide.

Next, set up the following table to show the equilibrium concentrations of the ions:

 

[ ]

[]

Initial

0.00

0.00

Change

+S

+2S

Equilibrium

S

2S

 Now substitute in the values of the concentrations of the ions into the expression to find .

Now, plug in the given  value and solve for , the molar solubility.

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