1:34 (Triple only) understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm³) of solution

Concentration is a measurement of the amount of substance per unit volume.

In Chemistry, concentration is measured in mol/dm³ (read as moles per cubic decimetre).

The following formula allows for the interconversion between a concentration (in mol/dm³), the amount (in moles) of a substance in a solution, and the volume of the solution (in dm³).

$amount(mol) = concentration(mol/dm^3) \times volume(dm^3)$

$concentration(mol/dm^3) = \frac{amount(mol)}{volume(dm^3)}$

Note: 1 dm³ = 1000 cm³

For example, to convert 250 cm³ into dm³:

$250cm^3 = \frac{250}{1000}$

$250cm^3 = 0.25dm^3$

Example 1: 0.03 mol of sodium carbonate (Na₂CO₃) is dissolved in 300 cm³ of water. Calculate the concentration of the solution.

Step 1: Convert the volume of water from cm³ into dm³.

$volume = 300cm^3$

$volume = \frac{300}{1000}dm^3$

$volume = 0.300dm^3$

Step 2: Use the molar concentration formula to calculate of the concentration of the solution.

$concentration = \frac{amount}{volume}$

$concentration = \frac{0.03}{0.300}$

$concentration = 0.1 mol/dm^3$

Example 2: Calculate the amount, in moles, of 25cm³ of hydrochloric acid (HCl) with a concentration of 2 mol/dm³.

Step 1: Convert the volume of HCl from cm³ into dm³.

$volume = 25cm^3$

$volume = \frac{25}{1000}dm^3$

$volume = 0.025dm^3$

Step 2: Rearrange the molar concentration formula to calculate of the amount, in moles of HCl.

$amount= concentration \times volume$

$amount = 2 \times 0.025$

$amount = 0.050 mol$

Titration calculations, example 1

The titration method that is used to prepare a soluble salt is also used to determine the concentration of an unknown solution.

For example, a titration problem will look like this:

A pupil carried out a titration to find the concentration of a solution of hydrochloric acid (HCl). She found that 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide solution (NaOH) required 23.50 cm³ of dilute hydrochloric acid for neutralisation. Calculate the concentration, in mol/dm³ of the acid.

The chemical equation for this reaction is:

NaOH(aq) + HCl(aq) –> NaCl(aq) + H₂O(l)

volume 25.0cm³ 23.50cm³

concentration 0.100mol/dm³

It can be useful, as shown above, to write the values of the volumes & concentration underneath the equation.

Step 1: Calculate the amount, in moles, of the solution that you know the values for both volume and concentration. In this case sodium hydroxide (NaOH).

First convert the volume of NaOH from cm³ into dm³.

$volume = 25cm^3$

$volume = \frac{25}{1000}dm^3$

$volume = 0.025dm^3$

Then rearrange the molar concentration formula to calculate of the amount, in moles of NaOH.

$amount= concentration \times volume$

$amount = 0.100 \times 0.025$

$amount = 0.0025 mol$

Step 2: Deduce the amount, in moles of the solution with the unknown concentration. In this case hydrochloric acid (HCl).

From the chemical equation the ratio of NaOH to HCl is 1:1

Therefore if you have 0.0025 mol of NaOH, this will react with 0.0025 mol of HCl.

Step 3: Calculate the concentration of the hydrochloric acid (HCl).

Convert the volume of HCl from cm³ into dm³.

$volume = 23.5cm^3$

$volume = \frac{23.5}{1000}dm^3$

$volume = 0.0235dm^3$

Use the molar concentration formula to calculate of the concentration of the HCl.

$concentration = \frac{amount}{volume}$

$concentration = \frac{0.0025}{0.0235}$

$concentration = 0.106 mol/dm^3$

Titration calculations, example 2

25.0 cm³ of sodium carbonate solution (Na₂CO₃) of unknown concentration was neutralised by 30.0 cm³ of 0.100 mol/dm³ nitric acid (HNO₃).

Na₂CO₃(aq) + 2HNO₃(aq) –> 2NaNO₃(aq) + CO₂(g) + H₂O(l)

Calculate the concentration, in mol/dm³ of the sodium carbonate solution.

Step 1: Calculate the amount, in moles, of nitric acid (HNO₃).

$volume = 30cm^3$

$volume = \frac{30}{1000}dm^3$

$volume = 0.030dm^3$

$amount= concentration \times volume$

$amount = 0.100 \times 0.030$

$amount = 0.0030 mol$

Step 2: Deduce the amount, in moles of sodium carbonate (Na₂CO₃).

Using the equation:

Na₂CO₃(aq) + 2HNO₃(aq) –> 2NaNO₃(aq) + CO₂(g) + H₂O(l)

Ratio Na₂CO₃:HNO₃ = 1:2

$amount=\frac{0.0030}{2} = 0.0015mol$

Step 3: Calculate the concentration of sodium carbonate (Na₂CO₃).

$volume = 25cm^3$

$volume = \frac{25}{1000}dm^3$

$volume = 0.025dm^3$

$concentration = \frac{amount}{volume}$

$concentration = \frac{0.0015}{0.025}$

$concentration = 0.06 mol/dm^3$

Converting mol/dm³ into g/dm³

Concentration can also be expressed in g/dm³ (grams per cubic decimetre).

Therefore mol/dm³ can be converted into g/dm³.

Example: Convert 0.06 mol/dm³ of sodium carbonate (Na₂CO₃) into g/dm³

Step 1: calculate the relative formula mass (M_r) of sodium carbonate (Na₂CO₃).

$M_r= (2 \times 23) + (1 \times 12) (3 \times 16) = 106$

Step 2: Recall the formula giving the relationship between mass, amount and formula mass.

$mass= amount(in\,moles) \times M_r$

therefore

$mass/dm^3= moles/dm^3 \times M_r$

$mass/dm^3= 0.06 \times 106$

$mass/dm^3= 6.36 g/dm^3$

Spec	Q'stion	Ans	MS
1:34	Jan '12 (2), Q 8 a	Answer	MkS
1:34	Jan '12 (2), Q 8 b	Answer	MkS
1:34	Jan '12 (2), Q 8 b	Answer	MkS
1:34	Jan '12 (2), Q 8 c	Answer	MkS
1:34	Jan '12 (2), Q 8 c	Answer	MkS
1:34	Jan '12 (2), Q 8 d	Answer	MkS
1:34	Jan '12 (2), Q 8 d	Answer	MkS
1:34	Jan '12 (2), Q 8 e	Answer	MkS
1:34	Jan '14 (1), Q 12 a ii	Answer	MkS
1:34	Jan '14 (1), Q 13 d	Answer	MkS
1:34	Jan '14 (2), Q 4 f i	Answer	MkS
1:34	Jan '14 (2), Q 4 f ii	Answer	MkS
1:34	Jan '14 (2), Q 4 f iii	Answer	MkS
1:34C	Jan '15 (1), Q 11 d i	Answer	MkS
1:34C	Jan '16 (1), Q 12 a ii	Answer	MkS
1:34C	Jan '16 (1), Q 15 b iv	Answer	MkS
1:34C	Jan '16 (2), Q 6 c i	Answer	MkS
1:34C	Jan '16 (2), Q 6 c iii	Answer	MkS
1:34C	Jan '17 (1), Q 9 a ii	Answer	MkS
1:34C	Jan '18 (1), Q 13 c i	Answer	MkS
1:34	May '10 (2), Q 6 d i	Answer	MkS
1:34	May '10 (2), Q 6 d ii	Answer	MkS
1:34	May '10 (2), Q 6 d iii	Answer	MkS
1:34	May '11 (2), Q 8 d i	Answer	MkS
1:34	May '11 (2), Q 8 d ii	Answer	MkS
1:34	May '11 (2), Q 8 d iii	Answer	MkS
1:34	May '11 (2), Q 8 d iv	Answer	MkS
1:34	May '13 (1), Q 9 a i	Answer	MkS
1:34	May '13 (1), Q 9 a ii	Answer	MkS
1:34	May '13 (1), Q 9 d i	Answer	MkS
1:34	May '13 (1), Q 9 d ii	Answer	MkS
1:34	May '13 (1), Q 9 d iii	Answer	MkS
1:34	May '13 (1R), Q 9 a iii	Answer	MkS
1:34C	May '14 (1), Q 15 a ii	Answer	MkS
1:34C	May '14 (2R), Q 5 b ii	Answer	MkS
1:34C	May '15 (1), Q 8 f i	Answer	MkS
1:34C	May '15 (1), Q 8 f iii	Answer	MkS
1:34C	May '15 (1R), Q 15 d i	Answer	MkS
1:34C	May '15 (1R), Q 15 d ii	Answer	MkS
1:34C	May '15 (1R), Q 15 d iii	Answer	MkS
1:34C	May '15 (1R), Q 15 d iv	Answer	MkS
1:34C	May '15 (2R), Q 3 c	Answer	MkS
1:34C	May '15 (2R), Q 3 d	Answer	MkS
1:34C	May '16 (1R), Q 10 a i	Answer	MkS
1:34C	May '17 (1R), Q 15 b	Answer	MkS
1:34C	May '17 (1R), Q 15 c	Answer	MkS
1:34C	May '17 (2R), Q 8 a ii	Answer	MkS

1:34 (Triple only) understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm³) of solution

1:34 (Triple only) understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm³) of solution

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