11.
Molarity, volumes
and the concentration of solutions
Appendix on
How to make up a
standard solution is on a separate page
See also 14.3
dilution of solutions calculations

Revise section 7. moles and
mass before proceeding in this section 11 and eventually you may need to be
familiar with the use of the apparatus illustrated above, some of which give
great accuracy when dealing with solutions and some do not.

It is very useful to be know exactly how much of a dissolved
substance is present in a solution of particular concentration or volume of a
solution.
 So we need a standard way of comparing the concentrations of
solutions in.
 The more you dissolve in a given volume of
solvent, or the smaller the volume you dissolve a given amount of solute in,
the more concentrated the solution.

The concentration of an aqueous solution is usually
expressed in terms of moles of dissolved substance per cubic decimetre
(reminder mole formula triangle on the right),
 with units mol
dm^{3} (or mol/dm^{3})and this is called molarity, sometimes denoted in
shorthand as M.
 Note:
1dm^{3} = 1 litre = 1000ml = 1000 cm^{3}, so dividing
cm^{3}/1000 gives dm^{3}, which is handy to know since
most volumetric laboratory apparatus is calibrated in cm^{3} (or ml),
but solution concentrations are usually quoted in molarity, that is mol/dm^{3}
(mol/litre).
 Concentration is also expressed in a
'nonmolar' format of mass per volume e.g. g/dm^{3}
 Equal volumes of solution of the
same molar concentration contain the same number of moles of solute i.e.
the same number of particles as given by the chemical formula.

You need to be able to calculate
 the number of moles or mass of substance in an aqueous
solution of given volume and concentration
 the concentration of an aqueous solution given the
amount of substance and volume of water, for this you use the equation
.... (reminder molarity formula triangle on the right),

(1a)
molarity (concentration) of Z
= moles of Z / volume in dm^{3}
 This is sometimes referred to as the
moleconcentration,
 and you need to be able to rearrange
this equation ... therefore ...
 (1b) moles =
molarity (concentration) x volume in dm^{3} and
...
 (1c)
volume in dm^{3}
= moles / molarity (concentration)
 You may also need to know that ...
 (2)
molarity x formula mass of
solute = solute concentration in g/dm^{3}
 This is sometimes referred to as the
massconcentration,
 and dividing this by 1000 gives
the concentration in g/cm^{3}, and
 (3)
concentration in g/dm^{3}
/ formula mass = molarity in mol/dm^{3}
 both equations (2) and (3)
result from equations (1) and (4), work it out for yourself.
 and to sum up, by now you should
know:
 (4) moles Z
= mass Z / formula mass of Z
 (5)
1 mole = formula mass in
grams
 (6) molarity = moles/dm^{3}

Molarity
calculation Example 11.1

What mass of
sodium hydroxide (NaOH) is needed to make up 500 cm^{3} (0.500 dm^{3})
of a 0.500 mol dm^{3}
(0.5M) solution? [A_{r}'s: Na = 23, O = 16, H = 1]

1 mole of NaOH = 23 + 16 + 1 = 40g

molarity = moles / volume, so
mol needed = molarity x volume in dm^{3}

500 cm^{3} = 500/1000 =
0.50 dm^{3}

mol NaOH needed = 0.500 x 0.500 =
0.250 mol NaOH

therefore mass = mol x formula
mass

= 0.25 x 40 = 10g NaOH
required



Molarity
calculation Example 11.2

(a) How
many moles of H_{2}SO_{4} are there in 250cm^{3} of
a 0.800 mol dm^{3}
(0.8M) sulphuric acid solution?

(b)
What mass of acid is in this solution?
[A_{r}'s:
H = 1, S = 32, O = 16]

(a) molarity = moles /
volume in dm^{3}, rearranging equation for the sulfuric acid

(b) mass = moles x formula
mass

Molarity
calculation Example 11.3

5.95g of
potassium bromide was dissolved in 400cm^{3} of water. Calculate
its molarity. [A_{r}'s: K = 39, Br = 80]

moles = mass / formula
mass, (KBr = 39 + 80 = 119)

mol KBr = 5.95/119 = 0.050
mol

400 cm^{3} =
400/1000 = 0.400 dm^{3}

molarity = moles of
solute / volume of solution

molarity of KBr
solution = 0.050/0.400 = 0.125M



Molarity
calculation Example
11.4 This involves calculating concentration in other ways e.g.
mass/volume units

What
is the concentration of sodium chloride (NaCl) in g/dm^{3} and g/cm^{3}
in a 1.50 molar solution?

At. masses: Na = 23, Cl
= 35.5, formula mass NaCl = 23 + 35.5 = 58.5

since mass = mol x formula mass,
for 1 dm^{3}

concentration = 1.5 x 58.5 = 87.8 g/dm^{3}, and

concentration =
87.75 / 1000 = 0.0878 g/cm^{3}



Molarity
calculation
Example 11.5

A solution of calcium
sulphate (CaSO_{4}) contained 0.500g dissolved in 2.00 dm^{3} of water.

Calculate the concentration in (a) g/dm^{3}, (b) g/cm^{3}
and (c) mol/dm^{3}.

(a) concentration = 0.500/2.00
= 0.250 g/dm^{3}, then since 1dm^{3}
= 1000 cm^{3}

(b) concentration = 0.250/1000
= 0.00025 g/cm^{3}

(c) At. masses:
Ca = 40, S = 32, O = 64, f. mass CaSO_{4} = 40 + 32 + (4 x 16) = 136

Molarity calculation Example 11.6

If 5.00g of sodium chloride is
dissolved in exactly 250 cm^{3} of water in a calibrated volumetric flask, what is the
molarity of the solution?

A_{r}(Na) = 23, A_{r}(Cl)
= 35.5, so M_{r}(NaCl) = 23 + 35.5 = 58.5

mole NaCl = 5.0/58.5 = 0.08547

volume = 250/1000 = 0.25 dm^{3}

Molarity = 0.08547/0.25 =
0.342 mol/dm^{3}



Molarity calculation Example
11.7
There are more questions
involving molarity in section 12. on
titrations
and
section 14.3 on
dilution calculations and
Appendix on
How to make up a
standard solution is on a separate page
APPENDIX on SOLUBILITY
How do you find out how soluble
a substance is in water?
Reminder: solute + solvent ==> solution
i.e. the solute is what dissolves, the
solvent is what dissolves it and the resulting homogeneous mixture is the
solution.
The solubility of a substance is the maximum
amount of it that will dissolve in a given volume of solvent e.g. water.
The resulting solution is known as a saturated
solution, because no more solute will dissolve in the solvent.
Solubility can be measured and expressed in with
different concentration units e.g. g/100cm^{3}, g/dm^{3}
and molarity (mol/dm^{3}).
Solubility can also be expressed as mass of
solute per mass of water e.g. g/100g of water.
You can determine solubility by titration if the
solute reacts with a suitable reagent e.g. acid  alkali titration and it
is especially suitable for substances of quite low solubility in water e.g.
calcium hydroxide solution (alkaline limewater) can be titrated with standard
hydrochloric acid solution.
However, many substances like salts are very
soluble in water and a simple evaporation method will do which is described below
e.g. for a thermally stable salt like sodium chloride.
(1) A saturated solution is prepared by mixing
the salt with 25cm^{3} of water until no more dissolves at room
temperature.
(2) The solution is filtered to make sure no
undissolved salt crystals contaminate the saturated solution.
(3) Next, an evaporating dish (basin) is
accurately weighed. Then, accurately pipette 10 cm^{3} of the saturated
salt solution into the basin and reweigh the dish and contents.
By using a pipette, its possible to express
the solubility in two different units.
(4) The basin and solution are carefully heated
to evaporate the water.
(5) When you seem to have dry salt crystals, you
let the basin cool and reweigh it.
(6) The basin is then gently heated again and
then cooled and weighed again.
This is repeated until the weight of the dish and
salt is constant, proving that all the water is evaporated
By subtracting the original weight of the
dish from the final weight you get the mass of salt dissolved in the volume
or mass of saturated salt solution you started with.
You can repeat the experiment to obtain a
more accurate and reliable result.
(7) Calculations
By using a pipette it is possible to
calculate the solubility in two ways, expressed as two quite different
units.
Suppose the dish weighed 95.6g.
With the 10.0 cm^{3} of salt
solution in weighed 107.7g
After evaporation of the water the dish
weighed 96.5g
Mass of 10.0 cm^{3}
salt solution = 107.7  95.6 = 12.1g
Mass of salt in 10 cm^{3}
of salt solution = 96.5  95.6 = 0.9g
Mass of water evaporated = 107.7 
96.5 = 11.2g
(a) Expressing the solubility in grams salt
per 100 g of water
From the mass data above 0.9g of salt
dissolved in 11.2g of water
Therefore Xg of salt dissolves in 100g of
water, X = 100 x 0.9 / 11.2 = 8.0
Therefore the solubility of
the salt = 8.0g/100g water
You can scale this up to 80.0g/1000g H_{2}O,
or calculate how much salt would dissolve in any given mass of water.
You can also express the solubility as g
salt/100g of solution.
0.9g salt is dissolved in 12.1g of
solution, Xg in 100g of solution
Therefore X = 100 x 0.9 / 12.1 = 7.4, so
solubility = 7.4g/100g solution
These calculations do not require
the original salt solution to be pipetted. You can just measure
out approximately 10cm^{3} of the salt solution with
10cm^{3} measuring cylinder, and do the experiment and
these calculations in the exactly the same way.
(b) However, if you know the exact volume of
salt solution and the mass dissolved in it, then you can calculate the
concentration in g/dm^{3}, and if you know the formula mass of the
salt, you can calculate the molarity of the solution.
From part (a) we have 0.9g of salt in
10.0 cm^{3}
Therefore Xg will dissolve in 1000cm^{3}
solution, X = 1000 x 0.9 / 10 = 90g/1000 cm^{3}
Solubility of salt = 90g/dm^{3}
Suppose the formula mass of the salt was
200, calculate the molarity of the saturated solution.
moles salt = mass / formula mass = 90/200
= 0.45 moles
Therefore solubility of saturated
salt solution in terms of molarity = 0.45 mol/dm^{3}
NOTE Solubility varies with temperature,
see
Gas and salt solubility
in water and solubility curves, and it usually (but not always)
increases with increase in temperature. So, in the experiment described
above, the temperature of the saturated solution should be noted, or perhaps
controlled to be saturated at 20^{o}C or 25^{o}C.
Selfassessment Quizzes
[msc] type in answer
Honly
or
multiple choice
Honly
See also
Advanced level GCEASA2
acidalkali titration calculation questions
OTHER CALCULATION PAGES

What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass

Calculating relative
formula/molecular mass of a compound or element molecule

Law of Conservation of Mass and simple reacting mass calculations

Composition by percentage mass of elements
in a compound

Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)

Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination

Introducing moles: The connection between moles, mass and formula mass  the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)

Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)

Moles and the molar volume of a gas, Avogadro's Law

Reacting gas volume
ratios, Avogadro's Law
and GayLussac's Law (ratio of gaseous
reactantsproducts)

Molarity, volumes and solution
concentrations (and diagrams of apparatus)
(this page)

How to do acidalkali
titration calculations, diagrams of apparatus, details of procedures

Electrolysis products calculations (negative cathode and positive anode products)

Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy

Energy transfers in physical/chemical changes,
exothermic/endothermic reactions

Gas calculations involving PVT relationships,
Boyle's and Charles Laws

Radioactivity & halflife calculations including
dating materials
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