1. m = Mr x n n = m/Mr record your answers to 2 d.p.
Work out the answers to the questions
Formulae square:
m = Mr x n n = m/Mr record your answers to 2 d.p.
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Week 3 Reacting amounts
Welcome: bags and coats to back / side
Drinks and food away please

2. The process…
Drawing theoretical amounts into practical work - titrations
Week 1 balancing equations
Week 2 the mole – relating equations to reacting ratios. Converting moles to masses etc.
This week: molarity: strength of solutions
Theory plus practical work – leads to assessment of both factors.

3. Week 3 learning aims:
Recall calculating Mr and converting moles to masses and visa versa (lots in the exam)
Calculate reacting masses from equations.
Make up a standard solution
News!

4. Molarity or concentration in moles per dm3
Molarity is a measure of concentration in solutions and is defined as the number of moles of a solute dissolved in 1 dm3 (1000 cm3 or 1 litre) of solution.
 Think of it like dissolving sugar in tea!
Units are mol dm-3 or M

5. 1. 0 M hydrochloric acid contains ……
1.0 M hydrochloric acid contains ……..moles HCl per dm3 (remember 1 dm3 is the same as 1 litre)
1 M solution contains …..mole of substance dissolved in 1 litre of water
2.0 M sodium hydroxide contains …….moles NaOH per dm3
2 M solution contains …… moles of substance dissolved in 1 litre of water
0.1 M sodium carbonate contains ……moles Na2CO3 per dm3
0.1 M solution contains …… mole of substance dissolved in 1 litre of water 

6. In summary
Remember molarity / concentration is expressed as moles/dm3 or mol dm-3 or M
If a solution has a concentration of x mol dm-3
It contains x moles of solute dissolved in 1 dm3 of water
1dm3 is the same as 1 litre

7. The concentration of a solution remains constant no matter what volume is taken.
Think of you sipping tea that has a set amount of sugar dissolved in it, the tea will taste the same even if you pour some away and then drink it.
However the actual number of moles of solute depends on the volume.
So 1 mole of sugar will make tea taste different if it is dissolved in 1 cup or in 1 litre.

8. Number of moles in a solution
n = M V remember volume must be in cm
Number of moles in a solution
solution
amount of solute (mol)
1 litre of 2M NaOH
1 moles of solute
2 dm3 2M NaOH 4
0.5 dm3 2 M NaOH 1
0.1 dm3 2 M NaOH
0.2
250 cm3 0.1M HCl
0.025
5 cm3 1.5 M HCl
0.0075
100 cm3 10 M H2SO4 1

9. When things get more complicated…say we don’t want a litre we want 200cm3
We can use the formula:
Number of moles = conc in mol dm-3 x volume in cm3
in a given volume
This can be given as n = M V
1000
There are 1000cm3 in 1 litre or 1 dm3
Read through the first page of the Molarity booklet and answer the questions.

10. Other formulae for conversions
Number of moles in = conc in mol dm-3 x volume in cm3
a given volume
n = M V
1000
Try question 1 a-e on page 3

11. 1) How many moles of each solute are contained in the following solutions?
(a) 400 cm3 of 0.2 M sodium hydroxide
n =
1000
0.2 x 400
= 0.08 mol
(b) 1.5 dm3 of 0.85 M sodium chloride
n =
1000
0.85 x 1500
= mol
(c) 16.4 cm3 of 2M hydrochloric acid
n =
1000
2 x 16.4
= mol
cont…

12. Stretch & challenge questions at the back of booklet.
1) How many moles of each solute are contained in the following solutions?
(d) 1000 cm3 of M silver nitrate
n =
1000
0.1 x 1000
= 0.1 mol
(e) 250 cm3 of 0.1 M potassium dichromate(VI)
n =
1000
0.1 x 250
= mol
Stretch & challenge questions at the back of booklet.

13. Look at question 2
What is the mass of the following dissolved in 250cm3 of 0.100M solution?
This is a two-parter question.
I first find out the number of moles for the substance, as I did before using the equation:
n = M V
1000
This tells me the moles, but not the mass….
We must then go on to the second part using an equation you have met before:
Mass = mister mole
m = Mr x n
If we multiple the number of moles by its Relative molecular mass we will be given the number of grams .

14. 2) What mass of each of the following is dissolved in 250 cm3 of 0
2) What mass of each of the following is dissolved in 250 cm3 of M solution?
n =
1000
0.1 x 250
= mol
m =
n x Mr
Mr = 170
a) Silver nitrate
m = x 170 = 4.25 g
b) Hydrochloric acid
Mr = 36.5
m = x 36.5 = g
c) Sulphuric acid
Mr = 98
m = x 98 = 2.45 g
cont…

15. 2) What mass of each of the following is dissolved in 250 cm3 of 0
2) What mass of each of the following is dissolved in 250 cm3 of M solution?
d) Sodium hydroxide
Mr = 40
m = x 40 = 1.0 g
e) Potassium manganate(VII)
Mr = 158
m = x 158 = 3.95 g

16. Break!

17. So far
We know how to work out the number of moles in an amount of solute.
Whether the amount is in grams (n = Mr /m)
Or as a solution of a given concentration
(n = M V/ 1000)
If we want to work our the number of grams of substance we can use m = Mr n
Or if its in solution, we can still work out the number of grams by first working out the number of moles present (n = M V /1000) then converting the moles to grams (m = Mr n) n

18. Let’s continue…
How do we work out a concentration in g dm-3 to molarity ( a concentration expressed in moles)
We can convert g dm-3 to mol dm-3
Using the formulae:
Molarity (mol dm-3) = concentration (g dm-3)
molar mass (g mol-1)
(The molar mass is the same as RMM but has units grams per mole)
M = c / Mr

19. 3) Calculate the molarity of
(a) 2 g of sodium hydroxide dissolved in 1 dm3 of solution
Molarity (M) =
Molar mass (Mr)
Concentation in g dm-3 (c)
C = 2 g dm-3
Mr = 40
M = 40 2
= 0.05 mol dm-3

20. 3) Calculate the molarity of
(b) 4 g of sodium hydroxide dissolved in 100 cm3 of solution
c = 4 x 10 = 40 g dm-3
Mr = 40
M = 40
= 1 mol dm-3
note we need to multiply up quantities so that we have the grams in a litre or decimeter cubed.

21. 3) Calculate the molarity of
(c) 4.9 g of H2SO4 dissolved in 250 cm3 of solution
c = 4.9 x 4 =19.6 g dm-3
Mr = 98
M = 98
19.6
= 0.2 mol dm-3

22. 3) Calculate the molarity of
(d) 5.3 g of Na2CO3 dissolved in 250 cm3 of solution
c = 4 x 5.3 =21.2 g dm-3.
Mr = 106
M =
106
21.2
= 0.2 mol dm-3

23. 3) Calculate the molarity of
(e) 25 g of CuSO4.5H2O dissolved in 1.5 dm3 of solution
c = 25 / 1.5 = g dm-3
Mr = 250
M =
250
16.67
If you are struggling with these hardly any in the exam.
= mol dm-3
Stretch and challenge at back

24. Titrations
When we have carried out practical work the past 2 weeks, we have carried out titrations.
This is where we use a solution of known concentration and volume (a standard solution) to work out the concentration of an unknown solution.
So far you have carried out your titration and found how much you have to add of the unknown solution to the set volume and concentration of your standard solution.
This value was your average titre. But we haven’t worked out it’s concentration, the objective of a titration.

25. This brings us to titration calculations
you did titrations where there was a reaction:
xA + yB - products
you carried out the titration a few times and got an average titre for the unknown concentration of solution.
You can use the following equation to work out its concentration.
MA VA = X
MB VB Y
MA = concentration in mol dm-3 of solution A in the equation
MB = concentration in mol dm-3 of solution B in the equation
VA = volume of solution A in the equation
VB= volume of solution B in the equation

26. Titration 1 HCl + NaOH  NaCl + H2O MAVA = MBVB 1 MAVA = MBVB MBVB MA=
23.5
0.1 x 25
= mol dm-3 (M)

27. Titration 2 H2SO4 + 2NaOH  Na2SO4 + 2H2O MBVB 2 = MAVA 1 MBVB = 2MAVA
24.3
2 x 0.1 x 25
= mol dm-3 (M)

28. Stretch and challenge more questions at the back
Titration 3
2HCl Na2CO  NaCl H2O CO2
(MAVA)
(MBVB)
Stretch and challenge more questions at the back
MBVB
MAVA = 1 2
MAVA = 2MBVB
MA= VA
2MBVB
MA =
19.4
2 x 0.05 x 25
= mol dm-3 (M)

29. Making a standard solution
This is a solution of known concentration.
This will be 250cm3 0.1M solution of sodium hydroxide
We need to know how many grams to weigh out.
Use the equation n = MV
1000
moles = 0.1 x 250 / 1000 = 0.025
Convert moles to mass …….m = Mr n
mass = ( ) x = 1g

30. Periodic Table
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