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Brown's Chemistry  GCSE/IGCSE/GCE (basic A level)
O Level
Online Chemical Calculations
11.
Introducing
Molarity, volumes
and the concentration of solutions
Help for problem solving in doing
molarity calculations from given masses, volumes and molecular/formula masses.
Practice revision questions on calculating molarity from mass, volume and
formula mass data, using experiment data, making predictions. How do we define the
concentration of a solution? How do we calculate concentration? What units do we
use for concentration? What is molarity? How do we use moles to calculate the
mass of a substance to make up a specific volume of a solution of specific
concentration? All calculation methods are fully explained with fully worked out example questions.
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11.
Molarity, volumes
and the concentration of solutions

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 gives
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}
 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}
 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 don't forget by now you should
know:
 (4) moles Z
= mass Z / formula mass of Z
 (5) 1 mole = formula mass in
grams
 (reminder molarity formula triangle
top right)

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}

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

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
There are more questions
involving molarity in section 12. on
titrations
and
section 14.3 on
dilution calculations
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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