I hope You all like it. I hope It is very beneficial for you all. I really thought that you all get enough knowledge from this presentation. This presentation is about materials and their classifications. After you read this presentation you knowledge is not as before.

1. INTERCONVERSION OF
CONCENTRATION UNITS SOLUTIONS

2. UNITS OF CONCENTRATION
 A solution is a homogeneous mixture of one
substance dissolved in another substance.
 Concentration is a ratio of the amount of
solute to the amount of solvent.

3. MOLARITY
 It is the number of moles of solute dissolved
per dm3 the solution.
 Molarity (M) = Number of moles of solute
Volume of solution (dm3)

4. MOLALITY
 It is the number of moles of solute in 1000g
of the solvent.
 Molality (m) = Number of moles of solute
Mass of solvent(kg)

5. PARTS PER MILLION (PPM)
 It is defined as the number of parts of solute per
million parts of the solution.
 Parts per million (ppm) = Mass of solute x106
Mass of solution

6. MOLE FRACTION
 The mole fraction of any component in a
mixture is the ratio of the number of moles of it
to the total number of moles of all the
components present.
 Mole fraction of any component can never be
greater then 1 and sum of all mole fractions is
equal to 1.
 xa = na
na + nb + nc

7. NORMALITY
 It is the equivalent weight of solute dissolved per liter
of solution.
 Normality = Equivalent weight of solute
1 liter of solution
 Equ.weight of solute = Molecular weight of solute
No. of replicable H+,OH- ions

8. PERCENTAGE COMPOSITION
I. Percentage Weight/Weight: It is the weight of
solute dissolved per 100 parts by weight of
solution.
II. Percentage Weight/Volume: It is the weight of
solute dissolved per 100 parts by volume of
solution.
III. Percentage Volume/Weight: It is the number of
cm3 solute dissolved per 100 g of the solution.
IV. Percentage Volume/Volume: It is the volume of
solute dissolved per 100 cm3 of the solution.

9. DILUTION OF THE SOLUTIONS
 Often we are given with solutions which are
concentrated more then needed so we have
to dilute those solutions.
 To determine how to dilute the stock
solution, use the formula:
C1V1 = C2V2
Where,
C1 = Concentration of stock
C2 = Concentration of diluted solution
V1 = Volume needed of stock
V2 = Final volume of dilution

10. CONVERSION FROM MOLARITY TO MOLALITY
 For converting a given concentration unit
from molarity to molality first assume that
you have 1 liter of solution.
 Find the mass of solution by multiplying
assumed volume with given density.
 Calculate grams of solute by given molarity
and find mass of solvent by difference of
mass of solution and mass of solute.
 Use values in molality formula to calculate
molality.

11. Problem 1: What is the molality of 20M H2SO4 given
that solution has a density of 2.65 g/mL?
 Volume of solution = 1 L = 1000 mL
Mass of solution = 1000 mL x 2.65 g/mL
= 2650g
20M solution contains H2SO4 = 20 x 98.08
= 1961.6g
Mass of solvent = 2650g – 1961.6g
= 688.4g
Molality = 20 moles = 29.0529 m H2SO4
0.6884 kg

12. CONVERSION FROM MOLALITY TO MOLARITY
 For converting a given concentration unit from
molality to molarity first assume that mass of
solvent is 1 kg.
 Find the mass of solute by given molality.
 Find mass of solution by adding mass of solute and
mass of solvent.
 Calculate the volume of solution by multiplying
given density with calculated mass of solution.
 Use calculated values in molarity formula for
calculation of molarity.

13. Problem 2: What is the molarity of 21.4 m HF
given that the solution has a density of 1.101
g/mL?
 Mass of solvent = 1 kg
Mass of HF (solute) = 21.4 x 20.01
= 428.21 g
Mass of solution = 1000g + 428.21 g
= 1428.21 g
Volume of solution = 1428.21g / 1.101g/mL
= 1297.1935 mL = 1.297 L
Molarity = 21.4 moles / 1.29719L
= 16.5 M HF

14. CONVERSION FROM PPM TO MOLARITY
 First of all find mass of solute from given ppm.
Taking weight of solution 1 L.
 Calculate no. of moles of solute by diving mass by
molecular mass.
 Use calculated values in molarity formula to find
molarity.

15. Problem 3: What is molarity of a dye
concentration given molar mass of dye 327
g/mol and of 2 ppm?
2 ppm solution = 2 mg/L
Mass of solute = 2 mg
Mass of solution = 1000 g
No. of moles of solute = 2 x 10-3g
327 g/mol
= 6.116 x 10-6 mol
Molarity = 6.116 x 10-6 mol/L or 6.116 x 10-6 M

16. CONVERSION FROM MOLARITY TO PPM
 Find mass of solute by given molarity.
 As is case of molarity , volume of solution is 1 L or
in case of mass 1000g.
 Put the calculated values in the formula of ppm to
find ppm concentration of given substance.

17. Problem 4: What is the ppm concentration of
calcium ion in 0.01M CaCO3?
 1 M CaCO3 contains Ca = 40 g
0.01 M CaCO3 contains Ca = 0.4 g
Mass of solution = 1000g
parts per million = 0.4 x 106
1000
= 400 ppm

18. CONVERSION BETWEEN MOLARITY AND PPM
 In general,
Normality = n-factor x Molarity
 Similarly:
Molarity = Normality / n-factor
Where n-factor means H+,OH- ions replaced by 1
mole of acid or base in a reaction.

19. Problem 5: What is the molarity of 1N H2SO4
solution?
 In this case n- factor = 2 (since there are 2 H+
ions)
Molarity = ½
= 0.5 M solution
Problem 6: What is the normality of A solution of
H2SO4 having a molarity of 0.100M?
 N-factor = 2 ( again 2 H+ ions replicable)
Nomality = 2 x 0.100
= 0.200 N solution

20. CONVERSION FROM MOLE FRACTION TO
MOLARITY
 Find out moles of solute and solvent by subtracting
given mole fraction from 1.
 Find mass of solute and mass of solvent by
multiplying moles with molar masses.
 Find mass of solution by adding mass of solute and
solvent.
 Calculate volume of solution by dividing its mass by
density.
 Use calculated values in molarity formula to
calculate it.

21. Problem 7: The mole fraction of an aqueous
solution of potassium oxide is 0.312. Calculate
the molarity (in mol/L) of the potassium oxide
solution, if the density of the solution is 1.28 g
mL-1.
 moles K2O = 0.312 mol
moles water = 1 - 0.312 = 0.688 mol
mass of water = 0.688mol x 18.02 g/mol
= 12.4 g
mass of K2O = 0.312mol x 94.199 g/mol
= 29.4 g
mass of solution = 41.8 g
volume solution = 41.8 / 1.28=32.7 mL
= 0.0327 L
Molarity = 0.312 mol / 0.0327 L = 9.55 mol/L
Or 9.55 m

22. CONVERSION FROM MOLARITY TO MOLE
FRACTION
 As in case of molarity , volume of solution is 1000ml
or in case of mass 1000g.
 Calculate moles of solvent, as in case of 1000g
water no. of moles are 55.5 moles.
 Find mole fraction of desired substance by using
mole fraction formula.

23. Problem 8: The molarity of HCl in water is 0.18.
What is its mole fraction?
 Moles of HCl in 1000 ml of water is 0.18.
Moles of water = 1000g
18g/mol
= 55.55 mol
Mole fraction = 0.18
0.18 + 55.55
= 0.0032

24. CONVERSION FROM % BY MASS TO
MOLARITY AND MOLALITY
 Find mass of solute and mass of solvent from given
percent.
 Find molar mass of solute and calculate its no. of
moles.
 Calculate volume of solute by dividing mass of
solution by given density.
 Use calculated values in molarity formula to
calculate it.
 In case of molality instead of solution, divide no. of
moles of solute by kg weight of solvent.

25. Problem 9: What is the molarity of the solution
that is 5.50% by mass oxalic acid
(C2H2O4), and has a density of 1.0224g/mL?
 Mass of C2H2O4 = 5.50g
Molar mass C2H2O4 = 90.36 g/mol
No. of moles of C2H2O4 = 5.50g / 90.36 g/mol
= 0.060867 moles
Volume of solution = 100g / 1.0224 g/mL
= 97.8090 mL = 0.0978L
Molarity = 0.060867mol / 0.0978L
= 0.6223M

26. IN THE SAME WAY, IF YOU ARE GIVEN
1 UNIT OF CONCENTRATION YOU CAN
CONVERT IT INTO REQUIRED UNIT BY
ABOVE METHODS.