|
 11.
Molarity, volumes
and the concentration of solutions
Selection of apparatus that might be associated with concentration
calculations e.g. involving molarity
See also 14.3
dilution of solutions calculations
(a)
Explaining
the terms solubility, concentration, strength and molarity
-
Why are the terms 'concentration', 'strength' and 'molarity' important?
-
Quite a lot of analytical
procedures in chemistry involve the use of solutions of accurately known
concentration e.g. standard solutions for various analytical purposes
including titrations.
-
If you want to
analyse an acid solution you need to titrate it with a standard solution of
alkali of accurately known concentration e.g. an accurately known molarity (concentration usually expressed in
mol/dm3,
lots more on this on the rest of this page!).
-
You can then do
a molarity calculation to ascertain the molarity of the unknown
concentration.
-
See
Acid-alkali
titration calculations, diagrams of apparatus, details of procedures
-
The solubility of a
substance is the maximum amount
of solute that dissolves in a given volume of solvent.
-
It is important
to know the solubility of substance in various liquids, quite often quoted
as the
maximum solubility of salts in water, but often quoted, not as molarity, but
in g salt /100 g of water and plotted in graphs known as solubility
curves (examples shown below) - based on the maximum concentration possible for a given solute and solvent
at specific temperatures.
- The diagrams represent two substances dissolved in the
solvent, the right-hand diagram represents a more concentrated solution.
- Most defined solutions contain only one substances, but others may
contain a mixture of several solutes depending on what the solution is used
for.
- The solution on the right has a greater molarity.
- The pictures do not mean the right hand one is a
stronger solution!
- Unfortunately, in everyday language, it would be
described as such, but this is science and the correct use of scientific
language is essential!
TOP OF PAGE
and INDEXES
OR
Quiz 11 on the basics of molarity and solution
concentrations
(b) Measures of concentration and simple calculations of
molarity
(b)(i) Concentration in terms of mass of
solute per unit volume of solution
It
is better to know how to rearrange an equation than use a formula triangle
A summary of how to do basic
concentration
calculations and rearrangement of the solution concentration formula
- We will look at moles and molarity in (b)(ii)
- The simplest measure of concentration is mass of
solute
per unit volume of solvent e.g.
- concentration = mass of solute / volume of
solvent
- Take 5.0 g of salt dissolved in 500 cm3
of water.
- The concentration can be expressed in several ways.
- concentration = 5.0/500 =
0.01 g/cm3
- 1 dm3 = 1000 cm3, so 500 cm3
= 500/1000 = 0.50 dm3
- concentration = 5.0/0.50 =
10.0 g/dm3
- For interconversion: g/cm3 x 1000 = g/dm3
AND g/dm3/1000 = g/dm3
- -
- Sometimes the general formula c = m/v is used
- c = concentration, m = mass, v = volume
- rearrangements: m = c x v and
v
= m/c
- -
-
Set A example questions (not using moles)
- AQ1
What is the concentration in g/dm3 if 6.0 g of salt is dissolved
in 150 cm3 of water?
- AQ2
Given a salt solution of concentration 16 g/dm3, what mass of
salt is in 40 cm3 of the solution?
- AQ3
Given 5.0 g of a salt, what volume of water in cm3, should it be
dissolved in to give a solution of concentration of 12.5 g/dm3?
- Its also good to be able to do dilution'
calculations
in section 14.3
dilution of solutions
TOP OF PAGE
and INDEXES
OR
Quiz 11 on the basics of molarity and solution
concentrations
(b)(ii)
Concentration in terms of moles of solute per unit volume of solution i.e.
MOLARITY
A summary of how to do basic molarity
calculations and rearrangement of the molarity formula AND it is much better to be
able to rearrange the molarity equation than using a triangle.
-
 For
most analytical and calculation purposes the concentration of an aqueous solution is usually
expressed in terms of moles of dissolved substance per cubic decimetre of
solution
(reminder mole formula triangle on the right). 1 cubic decimetre (dm3)
= 1 litre (l) in old money!
- concentration =
molarity = moles of solute /
volume of solvent in dm3 (litres)
- Make sure you know how to calculate moles, see
the triangle on the right!
- Using concentration units of mol
dm-3 (or mol/dm3), the concentration is called molarity, sometimes denoted in
shorthand as M (old money again, take care!) and the word molar
is used too.
- Note: 1dm3 = 1 litre = 1000 ml = 1000 cm3, so dividing
cm3/1000 gives dm3, which is handy to know since
most volumetric laboratory apparatus is calibrated in cm3 (or ml),
but solution concentrations are usually quoted in molarity, that is mol/dm3
(mol/litre).
- Concentration is also expressed in a
'non-molar' format of mass per volume e.g. g/dm3
- You need to know all about moles to proceed
further on this page and get into 'molarity' ...
- ... so read
section 7. on moles and
mass - essential pre-reading for section 11 ...
- AND, if you can't understand molarity, you cannot
do titration calculations either!
- 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 use in
defining a specific molarity.
- A note on solutions of ionic compounds e.g.
- A 1.0 molar solution of magnesium chloride MgCl2(aq),
contains 1.0 mol/dm3 of magnesium ions (Mg2+), BUT 2.0
mol/dm3 in terms of the chloride ion (Cl-)
concentration.
- 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), so, for a
substance Z ...
-
(1)
molarity (concentration) of Z
= moles of Z / volume in dm3
- This is sometimes referred to as the molar concentration
(mole-concentration),
- and you need to be able to rearrange
this equation ... therefore ...
- (1b) moles =
molarity (concentration) x volume in dm3 and
...
- (1c)
volume in dm3
= moles / molarity (concentration)
- You can use the triangle on the right to help you
rearrange the equation for the basic definition of molarity, BUT it
is much better to know how to rearrange the equation:
-
molarity = moles ÷ volume (dm3)
- You may also need to know that ...
- (2)
molarity x formula mass of
solute = solute concentration in g/dm3
- This is sometimes referred to as the mass-concentration,
- and dividing this by 1000 gives
the concentration in g/cm3, and
-
 (3)
concentration in g/dm3
/ formula mass = molarity in mol/dm3
- 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/dm3
-
Set B questions
on molarity
-
Molarity calculation question BQ1
-
If 5.00g of sodium chloride is
dissolved in exactly 250 cm3 of water in a calibrated volumetric flask,
-
(a) what is the concentration in g/dm3?
-
(b) What is the
molarity of the solution?
-
Molarity calculation question BQ2
-
5.95g of
potassium bromide was dissolved in 400cm3 of water.
-
(a)
Calculate
the molarity of the solution.
-
[Ar's: K = 39, Br = 80]
-
(b) What is the concentration in grams
per dm3?
-
Molarity calculation question BQ3
-
Molarity calculation question BQ4
-
Molarity calculation question BQ5
-
Molarity calculation question BQ6
-
A solution of calcium
sulphate (CaSO4) contained 0.500g dissolved in 2.00 dm3 of water.
-
Calculate the concentration in (a) g/dm3, (b) g/cm3
and (c) mol/dm3.
-
Molarity
calculation question BQ7
- Samples of human blood plasma and urine contain about 0.15 g of sodium
ions per 50 cm3.
- Atomic mass of sodium Na or Na+ = 23.0
- (a) Calculate the molarity of sodium ions in these fluids.
- (b) In the biochemistry of medicine concentrations are of often
expressed as millimol/dm3 (millimoles per litre).
- Calculate the sodium ion concentration in millimol/dm3
- -
- (c) If a sample of blood plasma has a concentration of 140 millimoles/dm3
what is the concentration of sodium ions in g/100 cm3?
-
Worked out answers to set B questions on molarity
- It is also good to be able to do dilution'
calculations
There are more questions
involving molarity in section 12. on
titrations
and
section 14.3 on
dilution calculations
TOP OF PAGE
and INDEXES
OR
Quiz 11 on the basics of molarity and solution
concentrations
(c) SOLUBILITY
and concentration calculations
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/100cm3, g/dm3
and molarity (mol/dm3).
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 25cm3 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 cm3 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 cm3 of salt
solution in weighed 107.7g
After evaporation of the water the dish
weighed 96.5g
Mass of 10.0 cm3 salt solution = 107.7 - 95.6 =
12.1g
Mass of salt in 10 cm3
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 X g 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 H2O,
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, X g 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 10cm3 of the salt solution with
10cm3 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/dm3, 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 cm3
Therefore X g will dissolve in 1000cm3
solution, X = 1000 x 0.9 / 10 = 90g/1000 cm3
Solubility of salt = 90g/dm3
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/dm3
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 20oC or 25oC.
TOP OF PAGE
and INDEXES
OR
Quiz 11 on the basics of molarity and solution
concentrations
(d) How to make up a standard solution - a solution of precisely known
concentration
The method and procedure of how to make up a
standard solution of a soluble solid e.g. a salt, is fully described.
Procedure
for
making up a standard solution of known molarity
The method and procedure of how to make up a
standard solution of a soluble solid e.g. a salt, is fully described
Example 1.
Suppose you want to make up 250 cm3 of a salt
solution of concentration 20g/dm3 (20g/litre, 20g/1000cm3,
20g/1000ml).
c = m / v, m = c x v, m = 20 x 250 / 1000 =
5g
so 5g of the salt is needed to be made up into an
aqueous solution of exactly 250.0 cm3.
The procedure
 to
 is
described in detail after example 2. below.
Example 2.
To prepare a solution of known molarity, you need to work backwards from the
volume required and the molarity to see how much solid you need.
Suppose you want to make up 250 cm3 of a sodium chloride solution of concentration
0.20 moldm-3
Preliminary calculation:
From molarity
formula (on the right): moles = molarity (mol/dm3) x
volume (dm3)
and volume in cm3 / 1000 = dm3
moles NaCl needed = 0.20 x 250/1000 = 0.20 x 0.25 = 0.05 mol NaCl
Atomic masses: Na =23 and Cl = 35.5, so molar mass of NaCl = 23 + 35.5 = 58.5
From basic mole formula: mass of NaCl needed = mol NaCl x formula mass NaCl
mass of NaCl needed = 0.05 x 58.5 = 2.925 g (which is ok if you
have a 3 decimal place balance!), so
2.295g of pure NaCl salt is needed to made up 250.0 cm3 of
solution with a precise concentration of 0.20 mol/dm3.
Procedure to make the standard solution i.e. one of known
concentration of solid (in this case)
An
accurate one pan electronic balanced is set to zero (preferably with an accuracy
of two decimal places). A beaker is placed on the balance and the reading noted
(ignore the figures on the diagram).
 Very
carefully, with a spatula (not shown), salt crystals are added to the beaker
until it weighs exactly 2.925 grams more than the beaker. This can be a very fiddly
procedure if you want exactly 2.925g of salt.
 Pure
water (distilled/deionised) is then added to the beaker to completely dissolve the salt and use of a
stirring rod helps to speed up the process.
The amount of water you add to the
beaker should be much less than 250cm3 to allow for the transfer and
rinsing of the solution into the standard volumetric flask using a
'squeezy' wash bottle!
 Eventually
a clear solution of the salt should be seen, there should be no residual salt
crystals at the bottom of the beaker or on the sides of the beaker.
You can use
the wash bottle to rinse down any crystals on the side of the beaker, but watch
the volume you use..
 An
accurately calibrated 250cm3 volumetric flask should be washed out
and cleaned several times with pure water.
Then, the whole of the solution in
the beaker is transferred into the flask with the help of a funnel to avoid the
risk of spillage.
To make sure every drop of the salt solution ends up in the
flask, a wash bottle of pure water is used to rinse out the beaker several
times, AND rinse the stirring rod and the funnel too.
This is to ensure nothing
is lost in the transfer fro beaker to flask.
Then,
very carefully, the flask is topped up with pure water so the meniscus rests
exactly on the 250.0cm3 calibration mark, a teat pipette is useful
for the last few drops of water.
The stopper is placed on and the flask carefully
shaken quite a few times to ensure the salt solution is completely mixed up.
Finally, check the meniscus lies on the calibration mark, in case another
few drops are needed.
Either way, the last drops of water should be added most
carefully with a teat pipette.
Job done!
Note
on standard solutions of acids and alkalis
You can purchase standard solutions ready for use.
OR, a phial of concentrated acid or alkali, which you
dilute into a specified volume to give a specific molarity.
Apart from weighing out a solid, the procedure is the
same as
and
,
ensuring every drop from the phial is rinsed down the funnel into the
calibrated volumetric flask.
See dilution' calculations
in section 14.3
dilution of solutions
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OR
Quiz 11 on the basics of molarity and solution
concentrations
(e) Self-assessment Quizzes
on molarity calculations:
Quiz 11 on the basics of molarity and solution
concentrations
type in titration answer
QUIZ
or
multiple choice
titration
QUIZ
(good revision for A level students as
well as GCSE level students)
See also
Advanced level
acid-alkali titration calculation questions
TOP OF PAGE
and INDEXES
OR
Quiz 11 on the basics of molarity and solution
concentrations
(f) Learning objectives for
molarity calculations
Know the different units to express the concentration of a solution.
e.g.
g/dm3, g/cm3, mol/dm3
or g dm-3, g cm-3, mol dm-3
Know how to convert between moles and mass: moles = mass in g / relative
formula mass
Know the definition and equation for molarity = moles / volume in dm3
and be able to rearrange the molarity equation.
Be able to define and explain
and use
the terms solute, solvent, solubility, concentration, strength and molarity.
Know what a solubility curve is and how to read it.
Know what a standard solution is including the concept of a solution of
specific molarity.
Be able to perform molarity calculations in different situations.
Be able to define a standard solution and no the method of how to prepare a
standard solution of known molarity.
Be able to describe and explain a method for determining the solubility of a
salt in water e.g. in g/100 cm3 or molarity (mol / dm3).
(i) Key
revision points on molarity
Molarity Revision
Notes (GCSE/IGCSE Chemistry)
Definition of molarity
-
Molarity (M)
= concentration of a solution expressed as moles of solute per litre
of solution.
- Formula:
M = n / V
- where n = moles of solute and V =
volume of solution in litres or dm3
- For molarity calculations, it is essential
you know how to calculate the number of moles of solute.
- moles (n) = mass (g) / relative
formula mass (Mr)
- Formula:
n = m / Mr
Key Calculation of
molarity
- Moles (n) from mass:
n = mass (g) / molar mass (Mr)
- Molarity (M) from moles (n) and
volume (V):
M = n moles / volume V
(V in litres or dm3, not cm3).
- Rearrangements:
-
n = M x V
-
V = n / M
-
Mass = M x V x Mr
(i.e. moles x relative molar mass)
Student Exam Tips
for molarity questions
- Always convert cm³ to dm³:
divide by 1000.
- Write balanced chemical equations
before calculating moles.
- Show all working—marks
are often awarded for method even if the final answer is wrong.
- Watch for unit traps:
examiners often mix g/dm³ and mol/dm³.
- In titration questions, carefully identify
which solution is in the burette and which is in the flask.
- Use significant figures
correctly—match the precision of the data given.
Typical student
misconceptions about molarity
- Confusing molarity (mol/dm³)
with mass concentration (g/dm³).
- Forgetting to convert volume units
(e.g., using 25 cm³ instead of 0.025 dm³) in
titration problems
too.
- Assuming molarity is the same as molality
(it isn’t—molality is moles per kg of solvent, not per litre of solution).
- Neglecting to use the balanced
equation to find mole ratios in
titration problems.
- Thinking molarity is independent of
temperature—actually, volume expansion can slightly affect concentration.
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Quiz 11 on the basics of molarity and solution
concentrations
(g) WHERE NEXT?
Index of all my online chemical calculation
quizzes
All my
GCSE/IGCSE/US grade 8-10 Level Chemistry Revision
notes
All my UK A/AS/IB/US grade 11-12 (honors) pre-university
Chemistry Notes
Spotted any careless error?
or request a type of
calculation not covered?
if so please
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Quiz 11 on the basics of molarity and solution
concentrations
This is a BIG
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| OTHER CALCULATION PAGES
1.
What is relative atomic mass, relative isotopic mass, calculating relative atomic mass
2.
Calculating relative
formula/molecular mass of a compound or element molecule
3.
Law of Conservation of Mass and simple reacting mass calculations
4. Composition by percentage mass of elements
in a compound
5. Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
6a. 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 (see calculations section 14.)
6b.
Reacting masses, concentration of solution and volumetric titration calculations
(NOT using moles)
7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)
8. Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
9.
Moles and the molar volume of a gas, Avogadro's Law
10. Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
reactants to products)
11. Molarity, volumes and solution
concentrations (and diagrams of apparatus)
(this page)
12.
How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
13. Electrolysis products calculations (negative cathode and positive anode products)
14. Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
14.1
% purity of a product 14.2a
% reaction yield
14.2b
atom economy 14.3
dilution of solutions
14.4
water of crystallisation
calculation
14.5
how
much of a reactant is needed? limiting reactant calculations
Energy transfers in physical/chemical changes,
exothermic/endothermic reactions
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
Radioactivity and half-life calculations including
dating materials
|
TOP OF PAGE
and INDEXES
Quiz 11 on the basics of molarity and solution
concentrations
|
(h) Worked out
ANSWERS to the self-test questions on molarity
Set A example questions on concentration ANSWERS (not using moles)
-
AQ1
What is the concentration in g/dm3 if 6.0 g of salt is dissolved
in 150 cm3 of water?
-
AQ2
Given a salt solution of concentration 16 g/dm3, what mass of
salt is in 40 cm3 of the solution?
-
Q3
Given 5.0 g of a salt, what volume of water in cm3, should it be
dissolved in to give a solution of concentration of 12.5 g/dm3?
-
c = m / v, rearranging gives v = m / c
-
v = 5.0 / 12.5 = 0.40 dm3
-
volume
of water needed = 1000 x 0.40 =
400 cm3
Its also good to be able to do dilution'
calculations
in section 14.3
dilution of solutions
-
Set B example questions on molarity concentration
-
Molarity calculation question BQ1
-
If 5.00g of sodium chloride is
dissolved in exactly 250 cm3 of water in a calibrated volumetric flask,
-
(a) what is the concentration in g/dm3?
-
(b) What is the
molarity of the solution?
-
Ar(Na) = 23, Ar(Cl)
= 35.5, so Mr(NaCl) = 23 + 35.5 = 58.5
-
mole NaCl = 5.0/58.5 = 0.08547
-
volume = 250/1000 = 0.25 dm3
-
molarity = mol of solute / volume
of solvent
-
Molarity = 0.08547/0.25 =
0.342 mol/dm3
-
-
-
Molarity calculation question BQ2
-
5.95g of
potassium bromide was dissolved in 400cm3 of water.
-
(a)
Calculate
its molarity. [Ar's: K = 39, Br = 80]
-
moles = mass / formula
mass, (KBr = 39 + 80 = 119)
-
mol KBr = 5.95/119 = 0.050
mol
-
400 cm3 =
400/1000 = 0.400 dm3
-
molarity = moles of
solute / volume of solution
-
molarity of KBr
solution = 0.050/0.400 =
0.125 mol/dm3
-
-
-
(b) What is the concentration in grams
per dm3?
-
concentration = mass / volume, the volume
= 400 / 1000 = 0.4 dm3
-
concentration = 5.95 / 0.4 =
14.9 g/dm3
-
-
-
Molarity calculation question BQ3
-
What mass of
sodium hydroxide (NaOH) is needed to make up 500 cm3 (0.500 dm3)
of a 0.500 mol dm-3
(0.5M) solution? [Ar'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 dm3
-
500 cm3 = 500/1000 =
0.50 dm3
-
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 question BQ4
-
(a) How
many moles of H2SO4 are there in 250 cm3 of
a 0.800 mol dm-3
(0.8M) sulphuric acid solution?
-
(b)
What mass of acid is in this solution?
[Ar's:
H = 1, S = 32, O = 16]
-
(a) molarity = moles /
volume in dm3, rearranging equation for the sulfuric acid
-
(b) mass = moles x formula
mass
-
Molarity calculation question BQ5
-
This involves calculating concentration in other ways e.g.
mass/volume units
-
What
is the concentration of sodium chloride (NaCl) in g/dm3 and g/cm3
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 dm3
-
concentration = 1.5 x 58.5 = 87.8 g/dm3, and
-
concentration =
87.75 / 1000 =
0.0878 g/cm3
-
-
-
Molarity calculation question BQ6
-
A solution of calcium
sulphate (CaSO4) contained 0.500g dissolved in 2.00 dm3 of water.
-
Calculate the concentration in (a) g/dm3, (b) g/cm3
and (c) mol/dm3.
-
(a) concentration = 0.500/2.00
=
0.250 g/dm3, then since 1dm3
= 1000 cm3
-
(b) concentration = 0.250/1000
=
0.00025 g/cm3 (or from 0.500/2000)
-
(c) At. masses:
Ca = 40, S = 32, O = 64, formula mass CaSO4 = 40 + 32 + (4 x 16) = 136
-
Molarity calculation question BQ7
-
Samples of human blood plasma and urine contain about 0.15 g
of sodium ions per 50 cm3.
-
Atomic mass of sodium Na or Na+ = 23.0
-
(a) Calculate the molarity of sodium ions in these fluids.
-
mol Na+ = 0.15 / 23 = 0.006522 mol
-
volume = 50 / 1000 = 0.05 dm3
-
molarity = 0.006522 / 0.05 =
0.13 mol/dm3
(2 sf)
-
-
-
(b) In the biochemistry of medicine concentrations are of
often expressed as millimol/dm3 (millimoles per litre).
-
Calculate the sodium ion concentration in millimol/dm3
-
milli means 1000th, so 1 mol = 1000 millimoles
-
There concentration of sodium ions = 1000 x 0.13 =
130 millimoles/dm3
-
-
-
(c) If a sample of blood plasma has a concentration of 140
millimoles/dm3 what is the concentration of sodium ions in g/100
cm3?
-
140 millimoles/dm3 = 0.140 mol/dm3 (÷
1000 = molarity)
-
mol = mass / atomic mass
-
mass = mol x atomic mass = 0.14 x 23 = 3.22 g per dm3
or 1000 cm3.
-
Therefore mass Na+ ions in 100 cm3 = 3.22 / 10 =
0.322 g/100 cm3
-
-
Its also good to be able to do dilution'
calculations
See section 14.3
dilution of solutions
Quiz 11 on the basics of molarity and solution
concentrations
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Quiz 11 on the basics of molarity and solution
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