Chemistry Reference
In-Depth Information
Moles and gases
Many substances exist as gases. If we want to find
the number of moles of a gas we can do this by
measuring the volume rather than the mass.
Chemists have shown by experiment that:
same number of molecules. This idea was also first
put forward by Amedeo Avogadro and is called
Avogadro's Law
.
Moles and solutions
Chemists often need to know the concentration
of a solution. Sometimes it is measured in grams
per cubic decimetre (g dm
−3
) but more often
concentration is measured in
moles per cubic
decimetre (mol dm
−3
)
. When 1 mole of a substance
is dissolved in water and the solution is made up to
1 dm
3
(1000 cm
3
), a
1 molar
(
1 mol dm
−3
) solution
is produced. Chemists do not always need to make
up such large volumes of solution. A simple method
of calculating the concentration is by using the
relationship:
One mole of any gas occupies a volume of
approximately 24 dm
3
(24 litres) at room
temperature and pressure (rtp). This quantity is also
known as the molar gas volume,
V
m
.
Therefore, it is relatively easy to convert volumes of
gases into moles and moles of gases into volumes
using the following relationship:
volume of the gas (in dm
3
at rtp)
24 dm
3
number of moles
of a gas
=
number of moles
volume (in dm
3
)
or
volume of
a gas
concentration (in mol dm
−3
) =
24 dm
3
(in dm
3
at rtp)
=
number of moles
of gas
×
Example 1
Calculate the concentration (in mol dm
−3
) of a
solution of sodium hydroxide, NaOH, which was
made by dissolving 10 g of solid sodium hydroxide
in water and making up to 250 cm
3
. (
A
r
: H = 1;
O = 16; Na = 23)
1 mole of NaOH contains 1 mole of sodium,
1 mole of oxygen and 1 mole of hydrogen.
Therefore:
mass of 1 mole of NaOH
= (1 × 23) + (1 × 16) + (1 × 1)
= 40 g
number of moles of NaOH in 10 g
mass of NaOH
mass of 1 mole of NaOH
Example 1
Calculate the number of moles of ammonia gas,
NH
3
, in a volume of 72 dm
3
of the gas measured
at rtp.
number of moles
of ammonia
volume of ammonia in dm
3
24 dm
3
=
72
24
=
= 3
Example 2
Calculate the volume of carbon dioxide gas, CO
2
,
occupied by
a
5 moles and
b
0.5 mole of the gas
measured at rtp.
=
10
40
=
a
volume of CO
2
= number of moles of CO
2
× 24 dm
3
= 5 × 24
= 120 dm
3
b
volume of CO
2
= number of moles of CO
2
× 24 dm
3
= 0.5 × 24
= 12 dm
3
The volume occupied by one mole of any gas must
contain 6 × 10
23
molecules. Therefore, it follows
that equal volumes of all gases measured at the
same temperature and pressure must contain the
= 0.25
(250 cm
3
=
250
1000
dm
3
= 0.25 dm
3
)
concentration of the NaOH solution
number of moles of NaOH
volume of solution (dm
3
)
=
0.25
0.25
=
= 1 mol dm
−3
