Chapter 6:  Solutions

 

 

Section 6-1:╩ Molarity

Section 6-2:╩ Solution Stoichiometry

Section 6-3:╩ Dilution

Chapter 6 Practice Exercises and Review Quizzes

 

 

 

 


  Section 6-1:  Molarity

 

A solution is a mixture of one substance, known as the solute, dissolved in a second substance, known as the solvent.  A common unit used to express the concentration of a solution is molarity (M), which is found by dividing the moles of solute by the total volume of the solution in liters:

 

 

Note that we do not actually need to know the volume or identity of the solvent to calculate molarity, as demonstrated in the following problem:

 

Sample Exercise 6A:

 

Calculate the molarity of a solution containing 7.6 grams of I2 dissolved to make 125 milliliters of solution.

 

Solution:

 

First, convert the mass of the solute I2 to moles and then divide by the volume of the solution in liters:

 

 

The molarity equation above can be rearranged to solve for either moles of solute or volume of solution:

 

 

When finding moles of solute or volume of solution, rather than plugging the measurements given in the problem into one of the two equations above, we will instead use dimensional analysis with molarity in mol/L as a conversion factor between moles of solute and volume of solution as demonstrated in the following two problems:

 

Sample Exercise 6B:

 

Calculate the mass of sucrose, C12H22O11, dissolved in 500. mL of 0.140 M solution.

 

Solution:

 

 

Sample Exercise 6C:

 

How many milliliters of 0.62 M solution contain 3.18 grams of dissolved C10H8?

 

Solution:

 

 

 

 

 

Section 6-2:  Solution Stoichiometry

 

For reactions involving at least one solution, we can use molarity to convert to and/or from moles in stoichiometry problems, as demonstrated below:

 

Sample Exercise 6D:

 

Given the unbalanced equation below, if 15 mL of 0.46 M KOH react, what mass of solid Mg(OH)2 will be produced?

 

Mg(NO3)2 (aq) + KOH (aq) Mg(OH)2 (s) + KNO3 (aq)

   

Solution:

 

For reactions involving at least one solution, molarity can also be used to convert to and/or from moles in limiting reagent and percent yield problems.

 

 

 

Section 6-3:  Dilution

 

When more solvent is added to dilute a solution, the final volume increases and the final molarity decreases, but the moles of solute do not change:

 

ni of solute = nf of solute

 

If we substitute the product of volume and molarity for moles in the equation above, we obtain an equation that is useful for many aspects of dilution (and also when solvent is removed through evaporation):

 

ViMi = VfMf

 

When using this dilution equation, note that the volume must be in the same unit on both sides of the equation but does not necessarily need to be changed to L and can remain in mL, as demonstrated in the following problem:

 

Sample Exercise 6E:

 

A student is given a bottle from the chemistry stockroom labeled 0.54 M NaCl.  If the student removes 12 mL of solution from the bottle and adds water until a volume of 36 mL is reached, what will be the new molarity of the solution?  

 

Solution:

 

Solve the dilution equation for final molarity:

 

 

 

 

Chapter 6 Practice Exercises and Review Quizzes:

 

6-1) Calculate the molarity of a solution containing 12.0 grams of C6H4Cl2 dissolved to make 975 milliliters of solution.

 

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6-1)

 

 

 

 

 

6-2) How many milliliters of 0.40 M solution contain 0.575 grams of dissolved Br2?

 

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6-2)

 

 

 

 

 

6-3) Given the unbalanced equation below, how many milliliters of 0.334 M HCl are needed to react with 3.82 grams of solid zinc?

 

Zn (s) + HCl (aq) ZnCl2 (aq) + H2 (g)

 

Click for Solution

 

6-3)

 

 

 

 

 

6-4) Given the unbalanced equation below, if 4.58 grams of solid CaCO3 is mixed with 75.0 mL of 0.645 M HI:

 

(a) Which is the limiting reagent?

(b) What is the maximum milliliters of carbon dioxide gas that can form at 35íC and 712 mmHg?

(c) What mass of the excess reagent remains when the reaction is complete?

 

CaCO3 (s) + HI (aq) CaI2 (aq) + H2O (l) + CO2 (g)

 

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6-4)

 

 

 

 

(a) HI produces less carbon dioxide, so HI is the limiting reagent.

 

(b)

 

 

 

 

 (c)

 

 

4.58 g – 2.42 g = 2.16 g CaCO3 excess

 

 

 

 

6-5) Given the unbalanced equation below, if 85 mL of 0.162 M AgNO3 reacts with an excess of Na2CO3 solution and then 1.8 grams of Ag2CO3 is actually collected, what is the percent yield of the reaction?

 

AgNO3 (aq) + Na2CO3 (aq) Ag2CO3 (s) + NaNO3 (aq)

 

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6-5)

 

 

 

 

6-6) You are in charge of providing 750. mL of 0.113 M HCl for your class to use in an experiment.  If you are given a bottle from the chemistry stockroom labeled 12.1 M HCl, how many milliliters of this more concentrated solution must you take out of the bottle and dilute with water in order to provide the correct volume and molarity of HCl solution for your class?

 

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6-6) Solve the dilution equation for initial volume:

 

 

 

 

 

 

 

 

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