Concentration of Solutions

Lessons

Senior 3 · Chemistry

Concentration of Solutions

5 lessons

Lesson 1 of 5

Molarity and Mass Concentration

25 min

Same sugar, different sweetness

Stir two heaped spoonfuls of sugar, about 10 g, into a 250 cm³ cup of tea, and the same 10 g into a 2-litre jug. The amount of sugar is identical, but the cup tastes far sweeter: each cubic centimetre of it holds eight times as much sugar as the jug. What your tongue detects is not the amount of sugar but how much is packed into each unit of volume, its concentration. In chemistry, where substances react particle by particle, the amount is counted in moles, and the volume in cubic decimetres.

Definition

Concentration: the amount of solute in a fixed volume of solution. The solute is the substance dissolved; the solvent (usually water) is the liquid it dissolves in.

Molar concentration (molarity), $c$ : the amount of solute, in moles, in $1\,\mathrm{dm^3}$ of solution.

$c = \frac{n}{V} \qquad \text{unit: } \mathrm{mol\,dm^{-3}}$

where $n$ is the amount in moles and $V$ the volume of the solution in $\mathrm{dm^3}$ . A solution of $2\,\mathrm{mol\,dm^{-3}}$ is often labelled "2 M" (two molar). Example: 0.50 mol of sodium chloride made up to $1\,\mathrm{dm^3}$ of solution has $c = 0.50\,\mathrm{mol\,dm^{-3}}$ .

Moles per cubic decimetre

Figure 1.1 shows why volume matters. Both beakers hold 0.10 mol of solute, but in the smaller volume the particles are twice as crowded, so the concentration is twice as great.

Volumes in the laboratory are measured in $\mathrm{cm^3}$ , so they must be changed to $\mathrm{dm^3}$ before you use $c = \frac{n}{V}$ :

$1\,\mathrm{dm^3} = 1000\,\mathrm{cm^3} = 1\,\text{litre} \qquad V\,(\mathrm{dm^3}) = \frac{V\,(\mathrm{cm^3})}{1000}$

The three quantities are linked by a formula triangle (Figure 1.2). Cover the one you want, and the other two show how to find it.

Worked example

Problem. 4.0 g of sodium hydroxide is dissolved in water and the solution is made up to $500\,\mathrm{cm^3}$ . Calculate its concentration in $\mathrm{mol\,dm^{-3}}$ . (Na = 23, O = 16, H = 1)

Solution. $M(\mathrm{NaOH}) = 23 + 16 + 1 = 40\,\mathrm{g\,mol^{-1}}$

$n = \frac{m}{M} = \frac{4.0}{40} = 0.10\,\mathrm{mol} \qquad V = \frac{500}{1000} = 0.500\,\mathrm{dm^3}$

$c = \frac{n}{V} = \frac{0.10}{0.500} = 0.20\,\mathrm{mol\,dm^{-3}}$

Worked example

Problem. What amount, and what mass, of sodium chloride is in $250\,\mathrm{cm^3}$ of a $0.40\,\mathrm{mol\,dm^{-3}}$ solution? (Na = 23, Cl = 35.5)

Solution. $n = c \times V = 0.40 \times \frac{250}{1000} = 0.10\,\mathrm{mol}$ . $M(\mathrm{NaCl}) = 58.5\,\mathrm{g\,mol^{-1}}$ , so

$m = n \times M = 0.10 \times 58.5 = 5.85\,\mathrm{g}$

Definition

Mass concentration: the mass of solute, in grams, in $1\,\mathrm{dm^3}$ of solution.

$\text{mass concentration} = \frac{m}{V} \qquad \text{unit: } \mathrm{g\,dm^{-3}}$

It is linked to molarity by the molar mass: $\text{mass concentration} = c \times M$ . Example: the sodium hydroxide solution above holds 4.0 g in $0.500\,\mathrm{dm^3}$ , so its mass concentration is $\frac{4.0}{0.500} = 8.0\,\mathrm{g\,dm^{-3}}$ , and $0.20 \times 40 = 8.0$ agrees.

Converting between g dm⁻³ and mol dm⁻³

$\text{g}\,\mathrm{dm^{-3}} \xrightarrow{\;\div M\;} \mathrm{mol\,dm^{-3}} \qquad \mathrm{mol\,dm^{-3}} \xrightarrow{\;\times M\;} \text{g}\,\mathrm{dm^{-3}}$

Table 1.1: The same solutions in both units

Solution	Molar mass (g mol⁻¹)	Molarity (mol dm⁻³)	Mass concentration (g dm⁻³)
sodium hydroxide, NaOH	40	0.20	8.0
sulfuric acid, $\mathrm{H_2SO_4}$	98	0.10	9.8
sodium carbonate, $\mathrm{Na_2CO_3}$	106	0.050	5.3
hydrochloric acid, HCl	36.5	2.0	73

Some solutes give more than one ion per formula unit. In $0.10\,\mathrm{mol\,dm^{-3}}$ calcium chloride, $\mathrm{CaCl_2(aq) \rightarrow Ca^{2+}(aq) + 2Cl^-(aq)}$ , the chloride ions have a concentration of $0.20\,\mathrm{mol\,dm^{-3}}$ .

Worked example

Problem. A bottle of sulfuric acid is labelled $24.5\,\mathrm{g\,dm^{-3}}$ . (a) What is its molarity? (b) What volume of it contains 0.050 mol of $\mathrm{H_2SO_4}$ ? (H = 1, S = 32, O = 16)

Solution. (a) $M(\mathrm{H_2SO_4}) = 2 + 32 + 64 = 98\,\mathrm{g\,mol^{-1}}$ , so $c = \frac{24.5}{98} = 0.25\,\mathrm{mol\,dm^{-3}}$ .

(b) $V = \frac{n}{c} = \frac{0.050}{0.25} = 0.20\,\mathrm{dm^3} = 200\,\mathrm{cm^3}$ .

Warning

Common mistakes. (1) Using $\mathrm{cm^3}$ in $c = \frac{n}{V}$ : always divide by 1000 first. (2) Taking $V$ as the volume of water added. It is the volume of the final solution: the solid is dissolved and then the solution is made up to the volume. (3) Mixing up the units: $\mathrm{mol\,dm^{-3}}$ counts particles, $\mathrm{g\,dm^{-3}}$ weighs them.

Tip

"n equals c V." Say it as one word, and it fits every concentration problem: $n = c V$ , with $V$ in $\mathrm{dm^3}$ . Molar mass is the bridge between grams and moles.

FuseSchool: Calculating Molarity. Watch how the amount in moles and the volume in cubic decimetres combine to give the concentration, and check each step against the worked examples above.

✦

Practice

Define molar concentration and give its unit.
Convert (a) $250\,\mathrm{cm^3}$ , (b) $25.0\,\mathrm{cm^3}$ to $\mathrm{dm^3}$ .
0.30 mol of potassium nitrate is dissolved and made up to $600\,\mathrm{cm^3}$ . Find the concentration.
5.3 g of sodium carbonate, $\mathrm{Na_2CO_3}$ , is made up to $250\,\mathrm{cm^3}$ of solution. Find (a) the mass concentration, (b) the molarity. (Na = 23, C = 12, O = 16)
How many moles of hydrochloric acid are in $20.0\,\mathrm{cm^3}$ of a $0.150\,\mathrm{mol\,dm^{-3}}$ solution?
What mass of potassium hydroxide, KOH, is needed to make $200\,\mathrm{cm^3}$ of a $0.50\,\mathrm{mol\,dm^{-3}}$ solution? (K = 39, O = 16, H = 1)
What is the concentration of chloride ions in $0.25\,\mathrm{mol\,dm^{-3}}$ magnesium chloride, $\mathrm{MgCl_2}$ ?
Solution A contains 0.020 mol of glucose in $100\,\mathrm{cm^3}$ ; solution B contains 0.15 mol in $1.0\,\mathrm{dm^3}$ . Which is more concentrated? Show your working.

Show answersHide answers

The amount of solute, in moles, in $1\,\mathrm{dm^3}$ of solution; $\mathrm{mol\,dm^{-3}}$ .
(a) $0.250\,\mathrm{dm^3}$ ; (b) $0.0250\,\mathrm{dm^3}$ .
$\frac{0.30}{0.600} = 0.50\,\mathrm{mol\,dm^{-3}}$ .
(a) $\frac{5.3}{0.250} = 21.2\,\mathrm{g\,dm^{-3}}$ ; (b) $M = 106$ , so $\frac{21.2}{106} = 0.20\,\mathrm{mol\,dm^{-3}}$ .
$0.150 \times 0.0200 = 0.00300\,\mathrm{mol}$ .
$n = 0.50 \times 0.200 = 0.10\,\mathrm{mol}$ ; $M = 56$ ; $m = 0.10 \times 56 = 5.6\,\mathrm{g}$ .
Each $\mathrm{MgCl_2}$ gives two $\mathrm{Cl^-}$ : $2 \times 0.25 = 0.50\,\mathrm{mol\,dm^{-3}}$ .
A: $\frac{0.020}{0.100} = 0.20\,\mathrm{mol\,dm^{-3}}$ ; B: $0.15\,\mathrm{mol\,dm^{-3}}$ . A is more concentrated, although it holds less glucose in total.

Sources and credits

Photo: Dilution-grenadine.jpg, Thierry Le Ridant, licensed under CC BY-SA 2.5, via Wikimedia Commons.
Video: Calculating Molarity, FuseSchool, embedded from YouTube.
Written by Ganzaa. The diagrams, table, worked examples and Practice items were drawn and written by Ganzaa.

1Same sugar, different sweetness

2Moles per cubic decimetre

3Converting between g dm⁻³ and mol dm⁻³

Same sugar, different sweetness

Moles per cubic decimetre

Converting between g dm⁻³ and mol dm⁻³