General Chemistry

Concentration is used to designate the amount of solute in a given quantity of solution. Molarity (M) is the most common way of expressing the solution concentration you will see in this chapter.

It expresses the concentration of a solution as the number of moles of solute in a liter of solution. To determine the molarity, we divide the amount of solute in moles by the volume of solution in liters:

 

 

For example, if 0.50 mol KNO3 is dissolved in a 2.0 L solution, the molarity of the solution would be:

\[{\rm{M}}\;{\rm{ = }}\,\frac{{\rm{n}}}{{\rm{V}}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.50}}\;{\rm{mol}}}}{{{\rm{2}}{\rm{.0}}\;{\rm{L}}}}\; = \;0.25\;{\rm{mol/L}}\]

 

If the quantity of the solute is given in grams or other units, convert it to moles and then divide by the volume of the solution in liters.

For example, what is the molarity of the solution prepared by dissolving 33.5 g sodium oxalate (Na2C2O4) in 350.0 ml water?

  • First, determine the moles of sodium oxalate:

                                                           

\[{\rm{n}}\;\left( {{\rm{N}}{{\rm{a}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{33}}{\rm{.5}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{134}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.25}}\,{\rm{mol}}\]

 

  • Make sure the volume is in liters:

 

\[{\rm{V}}\;{\rm{ = }}\;{\rm{350}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;{\rm{mL}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.35}}\,{\rm{L}}\]

 

  • Use the formula for molarity:

 

\[{\rm{M}}\;{\rm{ = }}\,\frac{{\rm{n}}}{{\rm{V}}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.25}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.35}}\;{\rm{L}}}}\; = \;0.71\;{\rm{mol/L}}\]

 

Sometimes, the questions are reversed, and you may be asked to find the volume of the solution containing a given amount of the solute. For this, determine the moles of the solute first, and rearrange the formula of molarity to get an expression for the volume.

For example, how many liters of 0.850 M BaCl2 solution is needed to obtain 9.37 g of the salt?

 

  • First, covert the mass of BaCl2 to moles:

 

\[{\rm{n}}\;\left( {{\rm{BaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{9}}{\rm{.37}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{208}}{\rm{.2}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0450}}\,{\rm{mol}}\]

 

  • Rearrange the formula for molarity to determine V:

 

\[{\rm{V}}\;{\rm{ = }}\,\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.0450 mol}}}}{{{\rm{0}}{\rm{.850 M}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0529 L}}\]

 

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Practice

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1.

Calculate the molarity of each ion in the following solutions:

a) 0.150 M NaNO3
b) 0.250 M K2SO4
c) 0.280 M Ca(NO3)2
d) 0.350 M AlCl3

 

a)
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b)
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c)
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d)
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2.

How many grams of calcium chloride, CaCl2, are needed to prepare 2.30 L of a 0.600 M solution?

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3.

Calculate the molarity of the solution prepared by dissolving 23.4 grams of NaCl in enough water to make a 500.0 mL solution. Consider the molar mass of NaCl to be 58.5 g/mol.

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4.

What is the molarity of the solution prepared by dissolving 33.5 g sodium oxalate (Na2C2O4) in 250.0 ml water?

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5.

Commercial concentrated hydrochloric acid is 36.0% by mass with a density of 1.2 g/mL. Calculate the molarity of this solution.

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6.

How many grams of copper(II) sulfate pentahydrate (CuSO4·5H2O) is needed to prepare 500. mL solution of 0.480 M CuSO4(aq)?

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7.

An ICU nurse is preparing for an intravenous administration of glucose (C6H12O6). How many mL of this solution will be needed to provide 6.50 mmol of glucose if the solution is labeled as 0.350 M?

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8.

How many milliliters of 0.850 M BaCl2 solution are needed to obtain 9.37 g of the salt?

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