Amount of substance
Molar Volumes of Gases
 Definition 1: Molar volume of gases
One mole of gas occupies 22,4 dm^{3} at standard temperature and pressure.
This applies to any gas that is at standard temperature and pressure. In grade 11 you will learn more about this and the gas laws.
Note:
Standard temperature and pressure (S.T.P.) is defined as a temperature of 273,15 K and a pressure of 0,986 atm.
For example, 2 mol of ${\text{H}}_{2}$ gas will occupy a volume of 44,8 dm^{3} at standard temperature and pressure (S.T.P.). and 67,2 dm^{3} of ammonia gas (${\text{NH}}_{3}$) contains 3 mol of ammonia.
Molar concentrations of liquids
A typical solution is made by dissolving some solid substance in a liquid. The amount of substance that is dissolved in a given volume of liquid is known as the concentration of the liquid. Mathematically, concentration (C) is defined as moles of solute (n) per unit volume (V) of solution.
For this equation, the units for volume are dm^{3} (which is equal to litres). Therefore, the unit of concentration is mol·dm^{−3}.
 Definition 2: Concentration
Concentration is a measure of the amount of solute that is dissolved in a given volume of liquid. It is measured in mol·dm^{−3}.
Example 1: Concentration calculations 1
Question
If 3,5 g of sodium hydroxide ($\text{NaOH}$) is dissolved in 2,5 dm^{3} of water, what is the concentration of the solution in mol·dm^{−3}?
Answer
Find the number of moles of sodium hydroxide
Calculate the concentration
The concentration of the solution is 0,035 mol·dm^{−3}.
Example 2: Concentration calculations 2
Question
You have a 1 dm^{3} container in which to prepare a solution of potassium permanganate (${\text{KMnO}}_{4}$). What mass of ${\text{KMnO}}_{4}$ is needed to make a solution with a concentration of 0,2 mol·dm^{−3}?
Answer
Calculate the number of moles
$\text{C}={\displaystyle \frac{\text{n}}{\text{V}}}$ therefore:
Find the mass
$\text{m}=\text{n}\times \text{M}=0,2\text{mol}\times 158\text{g}\xb7{\text{mol}}^{1}=31,6\text{g}$
The mass of ${\text{KMnO}}_{4}$ that is needed is 31,6 g.
Example 3: Concentration calculations 3
Question
How much sodium chloride (in g) will one need to prepare 500 cm^{3} of solution with a concentration of 0,01 mol·dm^{−3}?
Answer
Convert the given volume to the correct units
$\text{V}=500{\text{cm}}^{3}{\displaystyle \frac{1{\text{dm}}^{3}}{1000{\text{cm}}^{3}}}=0,5{\text{dm}}^{3}$
Find the number of moles
$\text{n}=\text{C}\times \text{V}=0,01\text{mol}\xb7{\text{dm}}^{3}\times 0,5{\text{dm}}^{3}=0,005\text{mol}$
Find the mass
$\text{m}=\text{n}\times \text{M}=0,005\text{mol}\times 58,45\text{g}\xb7{\text{mol}}^{1}=0,29\text{g}$
The mass of sodium chloride needed is 0,29 g
Exercise 1: Concentration of solutions
5,95 g of potassium bromide was dissolved in 400 dm^{3} of water. Calculate its concentration.
We use $C=\frac{n}{V}$ to calculate the concentration.
First we need to find n:
The molecular formula for potassium bromide is KBr. The molar mass of potassium bromide is:
$39,1+79,9=119\text{g}\cdot {\text{mol}}^{1}$
The number of moles is:
$n=\frac{m}{M}$ 
$n=\frac{5,95}{119}$ 
$n=0,05\text{mols}$ 
We need to also convert the volume to the correct units:
$\frac{400c{m}^{3}}{1000}=0,4d{m}^{3}$
Finally we work out the concentration:
$C=\frac{n}{V}$ 
$C=\frac{0,05}{0,4}$ 
$C=0,125\text{mol}\cdot {\text{dm}}^{3}$ 
100 g of sodium chloride ($\text{NaCl}$) is dissolved in 450 cm^{3} of water.

How many moles of $\text{NaCl}$ are present in solution?

What is the volume of water (in dm^{3})?

Calculate the concentration of the solution.
a) The molar mass of sodium chloride is:
$$
The number of moles of sodium chloride is:
$$ 
$$$$ 
$$$$ 
b) To convert fro$$m $$ t$$o $$ we divide by 1 000:
$$
c) The concentration is:
$$ 
$$$C=\frac{1,7}{0,45}$ 
$C=3.801919969\text{mol}\cdot {\text{dm}}^{3}$ 
What is the molarity of the solution formed by dissolving 80 g of sodium hydroxide ($\text{NaOH}$) in 500 cm^{3} of water?
The molar mass of sodium hydroxide is: $22,99+1,01+15,99=39,99\text{g}\cdot {\text{mol}}^{1}$.
The number of moles of sodium hydroxide is:
$n=\frac{m}{M}$ 
$n=\frac{80}{39,99}$ 
$n=2\text{mol}$ 
$\frac{500{\text{cm}}^{3}}{1000}=0,5{\text{dm}}^{3}$
The concentration is:
$C=\frac{n}{V}$ 
$C=\frac{2}{0,5}$ 
$C=4\text{mol}\cdot {\text{dm}}^{3}$ 
What mass (g) of hydrogen chloride ($\text{HCl}$) is needed to make up 1000 cm^{3} of a solution of concentration 1 mol·dm^{−3}?
$\frac{1000c{m}^{3}}{1000}=1d{m}^{3}$
The number of moles of hydrogen chloride is:
$n=CV$ 
$n=\left(1\right)\left(1\right)$ 
$n=1mol$ 
The molar mass of hydrogen chloride is:
$1,01+35,45=36,46g\cdot mo{l}^{1}$
The mass of hydrogen chloride needed is:
$m=nM$ 
$m=\left(1\right)(36,46)$ 
$m=36,46g$ 
How many moles of ${\text{H}}_{2}{\text{SO}}_{4}$ are there in 250 cm^{3} of a 0,8 mol·dm^{−3} sulphuric acid solution? What mass of acid is in this solution?
$\frac{250c{m}^{3}}{1000}=0,25d{m}^{3}$
The number of moles is:
$n=CV$ 
$n=(0,8)(0,25)$ 
$n=0,2mols$ 
The molar mass of ${H}_{2}S{O}_{4}$ is:
$2(1,01)+4(15,99)+32,07=98,05g\cdot mo{l}^{1}$
The mass of ${H}_{2}S{O}_{4}$ is:
$m=nM$ 
$m=(0,2)(98,05)$ 
$m=19,61g$ 