## 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

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)

0.150 M Naand 0.150 M NO3

b)

0.500 M K+ 0.250 M SO42-

c)

0.280 M Ca2+ and 0.560 M NO3

d)

0.350 M Al3+ 1.05 M  Cl

Solution

a) 0.150 M NaNO3

Each molecule/mole of NaNOcontains one/one mole of Naand one/one mole NO3 ion. Therefore, 0.150 M NaNOwill produce the same concentration of each ion.

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

NaNO3(aq) → Na+(aq) + NO3(aq)

Therefore, the concentration will be 0.150 M Naand 0.150 M NO3.

b) 0.250 M K2SO4

Each molecule/mole of K2SO4 contains two/two moles of Kand one/one mole SO42- ion. Therefore, 0.250 M K2SO4 will produce two times more K+ ions and the same concentration of SO42- ions.

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

K2SO4(aq) → 2K+(aq) + SO42-(aq)

Therefore, the concentration will be 0.500 M K+ 0.250 M SO42-.

c) 0.280 M Ca(NO3)2

Each molecule/mole of Ca(NO3)2 contains two/two moles of NO3 and one/one mole Ca2+ ion. Therefore, 0.280 M Ca(NO3)2 will produce two times more NO3 ions and the same concentration of Ca2+ ions.

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

Ca(NO3)2(aq) → Ca2+(aq) + 2NO3(aq)

Therefore, the concentration will be 0.280 M Ca2+ and 0.560 M NO3.

d) 0.350 M AlCl3

Each molecule/mole of AlCl3 contains three/three moles of Cl and one/one mole Al3+ ion. Therefore, 0.350 M AlCl3 will produce three times more Cl ions and the same concentration of Al3+ ions.

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

AlCl3(aq) → Al3+(aq) + 3Cl(aq)

Therefore, the concentration will be 0.350 M Al3+ 1.05 M  Cl.

2.

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

153 g

Solution

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.

0.800 mol/L

video

4.

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

1.0 mol/L

Solution

The molarity is calculated by the following formula:

M = n (mol)/V (L)

First, determines the moles of sodium oxalate:

n (Na2C2O4) = 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

5.

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

Solution

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. Ror 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

6.

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

59.9 g CuSO4·5H2O

Solution

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 CuSO4·5H2O. For this, rearrange the equation:

n (mol) = M x V (L) = 0.480 M x 0.500 L = 0.240 mol CuSO4

Now, these are the moles of CuSO4, but we are asked to find the needed mass of CuSO4·5H2O. Although the molar masses of these salts are different, their moles are the same because each mol of CuSO4·5H2O contains 1 mol CuSO4. Therefore, the moles of CuSO4·5H2O needed to prepare the solution are the same as the moles of CuSO4. In other words, if we need to supply 5 moles of CuSO4, then 5 moles of CuSO4·5H2O are going to be needed.
So, in this case, n(CuSO4·5H2O) = 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 CuSO4·5H2O

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?

19 mL

Solution

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

8.

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

52.9 mL

Solution

First, covert the mass of BaCl2 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