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 KNO_{3} 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 (Na_{2}C_{2}O_{4}) 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* BaCl_{2} solution is needed to obtain 9.37 g of the salt?

- First, covert the mass of BaCl
_{2}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:

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

**Check Also**

- Solutions
- Strong and Weak Electrolytes
- Dissociation of Ionic Compounds
- Molecular, Ionic, and Net Ionic Equations
- Dilution
- Ion Concentration
- Precipitation Reactions
- Definitions of Acids and Bases
- Acid-Base Reactions
- Stoichiometry of Reactions in Aqueous Solutions
- Acid-Base Titrations
- Oxidation State
- Oxidation-Reduction (Redox) Reactions
**Reactions in Aqueous Solutions Practice Problems**

#### Practice

Calculate the molarity of each ion in the following solutions:

a) 0.150 *M* NaNO_{3
}b) 0.250 *M* K_{2}SO_{4
}c) 0.280 *M* Ca(NO_{3})_{2
}d) 0.350 *M* AlCl_{3}

0.150 *M* Na^{+ }and 0.150 *M* NO_{3}^{–}

0.500 M K^{+}_{ }0.250 M SO4^{2-}

0.280 *M* Ca^{2+ }and 0.560 *M* NO_{3}^{–}

0.350 *M* Al^{3+ }1.05 *M* Cl^{–}

**a)** 0.150 *M* NaNO_{3}

Each molecule/mole of NaNO_{3 }contains one/one mole of Na^{+ }and one/one mole NO_{3}^{–} ion. Therefore, 0.150 *M* NaNO_{3 }will produce **the same concentration of each ion**.

You can also write the dissociation equation to see the mol/molar ratio:

NaNO_{3}(*aq*) → Na^{+}(*aq*) + NO_{3}^{–}(*aq*)

Therefore, the concentration will be 0.150 *M* Na^{+ }and 0.150 *M* NO_{3}^{–}.

_{
}**b)** 0.250 *M* K_{2}SO_{4}

Each molecule/mole of K_{2}SO_{4}_{ }contains two/two moles of K^{+ }and one/one mole SO_{4}^{2-} ion. Therefore, 0.250 *M* K_{2}SO_{4}_{ }will produce **two times more K ^{+ }**ions and the

**same concentration of SO**ions.

_{4}^{2- }You can also write the dissociation equation to see the mol/molar ratio:

K_{2}SO_{4}(*aq*) → 2K^{+}(*aq*) + SO_{4}^{2-}(*aq*)

Therefore, the concentration will be 0.500 M K^{+}_{ }0.250 M SO4^{2-}.

_{
}**c)** 0.280 *M* Ca(NO_{3})_{2}

Each molecule/mole of Ca(NO_{3})_{2}_{ }contains two/two moles of NO_{3}^{–}^{ }and one/one mole Ca^{2+} ion. Therefore, 0.280 *M* Ca(NO_{3})_{2}_{ }will produce **two times more NO _{3}^{–}^{ }**ions and the

**same concentration of Ca**ions.

^{2+}^{ }You can also write the dissociation equation to see the mol/molar ratio:

Ca(NO_{3})_{2}(*aq*) → Ca^{2+}(*aq*) + 2NO_{3}^{–}(*aq*)

Therefore, the concentration will be 0.280 *M* Ca^{2+ }and 0.560 *M* NO_{3}^{–}.

_{
}**d)** 0.350 *M* AlCl_{3}

Each molecule/mole of AlCl_{3}_{ }contains three/three moles of Cl^{–}^{ }and one/one mole Al^{3+} ion. Therefore, 0.350 *M* AlCl_{3}_{ }will produce **three times more Cl ^{–}^{ }**ions and the

**same concentration of Al**ions.

^{3+}^{ }You can also write the dissociation equation to see the mol/molar ratio:

AlCl_{3}(*aq*) → Al^{3+(}*aq*) + 3Cl^{–}(*aq*)

Therefore, the concentration will be 0.350 *M* Al^{3+ }1.05 *M* Cl^{–.}

How many grams of calcium chloride, CaCl_{2}, are needed to prepare 2.30 L of a 0.600 *M* solution?

153 g

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.

0.800 mol/L

What is the molarity of the solution prepared by dissolving 33.5 g sodium oxalate (Na_{2}C_{2}O_{4}) in 250.0 ml water?

1.0 mol/L

The molarity is calculated by the following formula:

M = n (mol)/V (L)

First, determines the moles of sodium oxalate:

n (Na_{2}C_{2}O_{4}) = 33.5 g / 134 g/mol = 0.25 mol

Next, convert the mL to L in order to match the formula.

V = 250 mL x 1 L/1000 mL = 0.25 L

And now, plug in the numbers to determine the concentration of sodium oxalate:

**M = n (mol)/V (L) = 0.25 mol/0.25 L = 1.0 mol/L**

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

11.9 mol/L

The molarity is calculated by the following formula:

M = n (mol)/V (L)

So, first, we need to determine the moles of hydrochloric acid. If the quantity of the solution is not given, you can assume any number. For convenience, 100 g solution is a good number. In the 100 g solution, there will be 36.0 g HCl. Therefore, the moles of HCl will be:

n (HCl) = m (g) / M (g/mol) = 36.0 g / 36.5 g/mol = 0.987 mol.

Next, we need to determine the volume of the solution which can be calculated using the mass and the density of the solution.

V = m (g) / d (g/mL) = 100.0 g / 1.2 g/mL = 83 mL

We also need to convert this to L:

V (L) = 83 mL x 1 L/1000 mL = 0.083 L

And last, the molarity will be:

M = n (mol)/V (L) = 0987 mol / 0.083 L = 11.9 mol/L

How many grams of copper(II) sulfate pentahydrate (CuSO_{4}·5H_{2}O) is needed to prepare 500. mL solution of 0.480 *M* CuSO_{4}(aq)?

**59.9 g CuSO _{4}·5H_{2}O**

The molarity is given by the following formula:

M = n (mol)/V (L)

We are given the molarity (0.480 M) and V (500. mL) and we need to find the mass. The link between the formula of molarity and mass is the moles. Therefore, we first need to calculate the moles of CuSO_{4}·5H_{2}O. For this, rearrange the equation:

n (mol) = M x V (L) = 0.480 M x 0.500 L = 0.240 mol CuSO_{4}

Now, these are the moles of CuSO_{4, }but we are asked to find the needed mass of CuSO_{4}·5H_{2}O. Although the molar masses of these salts are different, their moles are the same because each mol of CuSO_{4}·5H_{2}O contains 1 mol CuSO_{4. }Therefore_{, }the moles of CuSO_{4}·5H_{2}O needed to prepare the solution are the same as the moles of CuSO_{4}. In other words, if we need to supply 5 moles of CuSO_{4}, then 5 moles of CuSO_{4}·5H_{2}O are going to be needed.

So, in this case, n(CuSO_{4}·5H_{2}O) = n(CuSO4) = 0.240 mol

The mass is calculated from the equation

n (mol) = m (g) /M (g/mol)

therefore,

m (g) = n (mol) x M (g/mol) = 0.240 mol x 249.7 =** 59.9 g CuSO _{4}·5H_{2}O**

An ICU nurse is preparing for an intravenous administration of glucose (C_{6}H_{12}O_{6}). 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?

**19 mL**

Like the problems before, we need to use the formula for molarity. One difference here is that the amounts are in millimoles, so we need to do a conversion first.

Moles of glucose = 6.50 mmol x 1 mol/1000 mmol = 0.0065 mol

From here, we can find the volume of solution in liters and convert to mL.

V = n/M = 0.0065 mol/0.350 M = 0.019 L

V (mL) = 0.019 L x 1000 mL/1L = **19 mL**

How many milliliters of 0.850 M BaCl_{2} solution are needed to obtain 9.37 g of the salt?

**52.9 mL**

First, covert the mass of BaCl_{2} to moles.

n = m/M = 9.37 g / 208.2 g/mol = 0.0450 mol

Next, we can find the volume using the equation for molarity:

V = n/M = 0.0450 mol / 0.850 M = 0.0529 L

And now, convert the liters to milliliters:

V (mL) = 0.0529 L x 1000 mL/1L = **52.9 mL**