1. Different methods for concentration expression
Mole: The mole (mol) is a fundamental unit describing the amount of chemical species. It is always associated with the chemical formula and represented one Avogadro's number (6.02 x 1023) of atom, ions, molecules or elements.
No. of moles (A) = No. of (A) / Avogadro's number
No. of moles (A) = No. of (A) / 6.02 × 1023
No. of (A) = No. of moles (A) × × 1023
A= atoms, molecules, ions or electrons.
No. of moles = weight (g)/ Molecular weight (g/mol)
Weight (g) =No. of moles × Molecular weight (g/mol)

2. The Molecular weight (M.Wt):
Is the summation of the atomic weights of all the atoms appearing in the chemical formula of a substance.
e.g:
M.wt of methane CH4 =1× atomic weight of C + 4 × atomic weight of H
= 1 × × 1
= 16 g/mole
M.wt of ethanol C2H5OH = 2 × × × 16
= 46 g/mole

3. Millimole (mmol):
Some times it is more convenient to make calculations with millimoles rather than moles.
1 mol = 1000 mmol or 1 mmol = 1/1000 mol = 10-3 mol
No. of mmol = (weight (gm) / M.Wt.) x 1000

4. Ex 1: How many Na+ ions are contained in 5.0 moles of sodium ion?

5. Ex 2: How many moles of glucose (C6H12O6) are contained in 36 gm of the pure sugar?
If you know that: Atomic weight (At.wt): [H=1, C=12, O=16] g/mol

6. Ex 3: How many millimoles are contained in
a) 6.75 g of Al2O3?
b) 232 mg of Na2SO4?

7. Molarity [Molar concentration (M)]:
Number of moles of solute in one liter of solution.
M = No. of moles / volume(L) = No.of mmoles/volume (ml)
No. of moles = M × V (L) or No. of mmoles = M × V (ml)
No. of moles = Wt. of solute (g) /M.Wt (g/mole)

8. Ex : Calculate the molarity of a species in
a) An aqueous solution that contains 2.3 g of ethanol (C2H5OH) in 2.00L?
b) An aqueous solution that contains 285mg of trichloroacetic acid (Cl3CCOOH) in 10 ml?
Atomic weight (At.wt): [H=1, C=12, Cl=35.5, O=16] g/mole

9. Ex : An analyst wishes to add 120 mg NaOH base to dissolve an organic acid sample using a 0.248M NaOH solution. How many milliliters of NaOH solution should be added to the organic acid? At.wt: (Na =23, O =16, H =1) g/mol

10. Ex :
a) Calculate the molar concentration of potassium dichromate (K2Cr2O7) in a solution that contains 16.5 g/L (K2Cr2O7)?
b) How many moles of K2Cr2O7 will be present in 150 ml of the solution?
At.wt: (K=39, Cr=52, O=16) g/mol

11. No. of mole Wt(g)/M.Wt
M = = multiply by 1000/1000
V(L) V(L)
Wt(g) x 1000
M = Wt(g) x 1000 = M x V(ml) x M.Wt.
V(ml) x M.Wt
M x V(ml) x M.Wt
Wt(g) =
1000
This equation is very important in analytical chemistry especially in preparation of solutions from solid substances.

12. 1- Preparation of solutions from Solid compounds.
Ex:
Prepare 2 liter of 0.1M Na2CO3 from the solid material.
What is needed to solve this example…..?
Request: how much weighing from the solid material (Na2CO3)?
By another words :
How much gram we needed to weighing from Na2CO3 and dissolving in 2 liter of water to obtaining 0.1M Na2CO3?

13. Ans.
0.1M = 0.1 mole Na2CO3 in one liter of solution
= 0.1 x 2 = 0.2 mole Na2CO3 in 2liter of solution. (Because we want 2L)
Wt. (g)
no. of mole =
M.Wt.
Wt. (g) = no. of mole x M.Wt.
Wt. of Na2CO3 (g) = 0.2 x [(2x23) + (1x12) + (3x16)] = 21.2 g
which is required to dissolved in 2 liter of water to obtain 0.1M Na2CO3 in this volume (2L).

14. OR
M x V(ml) x M.Wt x 2000 x 106
Wt(g) = = = 21.2 g
(21.2 g) which is required to dissolved in 2 liter of water to obtain 0.1M Na2CO3 in this volume (2L)

15. Ex:
Prepare 0.1M of NaCl in 250ml volume?
Question requirement was how many grams of solid NaCl was required to prepare a solution of this salt in 250ml volume.
Ans:
0.1M = 0.1 mole NaCl in 1L of solution
= 0.1 x (250/1000) = mole NaCl in 250ml of solution.
Wt(g)
No. of mole = Wt = No.of mole x M.Wt
M.Wt
Wt. of NaCl (g) = x [(1x23) + (1x35.5)] = g NaCl
is required to dissolved in 250ml of water to obtain 0.1M NaCl.

16. OR
M x V(ml) x M.Wt x 250 x 58.5
Wt(g) = = = g
(1.4625g NaCl) is required to dissolved in 250ml of water to obtain 0.1M NaCl.

17. Q/ Explain how we can prepare each of the following solutions from pure solid substances:
1- Prepare 0.2M BaCl2.2H2O in 500ml.
2- Prepare 0.03M CaCl2 in 1L.
3- Prepare 1x10-3M CuSO4.5H2O in 100ml.
(At.Wt: Ba=137, Cl=35.5, H=1, O=16, Ca=40, Cu=63.5, S=32)

18. Ex:
Prepare 0.25M Cl- in 250ml from NaCl salt.

19. Ex:
Prepare 2 liter of 0.1M Na+ from Na2CO3 pure solid material.
What is needed to solve this example…..?
Request: how mush weighing (Na2CO3) by grams to dissolved in 2 liter to obtain solution contained 0.1M Na+?

20. Q/ Prepare 0.25M Cl- in 250ml from BaCl2.2H2O salt.

21. 2- Preparation of solutions from Liquids.
Preparation of solutions from unknown concentration of liquids.
This process was done by two steps:
The first step must be calculated molar concentration of concentrated solution (bottle).

22. Q/ Calculate H3PO4 molarity in the bottle from the information.
Specific gravity = 1.696
Percentage = 85%
Ans:
Must be convert (g/ml) to (mol/L) because we want molarity.
Density = g/ml
= x 1000 = 1696 g/L this is when the percentage of H3PO4 was 100%.
But the percentage was 85%
1696 x (85/100) = g/L for this solution (bottle).

23. Wt(g)/M.Wt g /M.Wt
M = =
V(L) L
1441.6
M = 98
= 14.7 mol/L (M) (molar concentration of the concentrated
acid (H3PO4) in the bottle).

24. There is another way to solve this example (fast way):
Sp.gr. x % x 1000
M =
M.Wt.
For the above example…..
1.696 x (85/100) x 1000
M = 98
= 14.7 mol/L (M) the same result.

25. The second step dilution process must be done by this relation:
(M1 V1)concentrated = (M2 V2)diluted
From first step unknown known

26. Ex: Prepare 500ml (6M) from the concentrated H3PO4 .
Sp.gr. = , percentage = %85 , At.Wt: P=31 , O=16.
Ans:
The first step must be calculated molar concentration of concentrated solution (bottle).

27. The second step: (dilution)
(M1V1)concentrated = (M2V2)diluted
(14.7 x V1) = (6 x 500)
V1 = (6x500)/14.7 = ml
This is meaning we take ml from the concentrated solution (bottle) and diluted to 500 ml with distilled water to obtain 6M H3PO4.

28. Ex: Prepare 250ml of 2M HNO3 from the concentrated solution
Ex: Prepare 250ml of 2M HNO3 from the concentrated solution. The information on the bottle:percentage = 69% , specific gravity 1.42, M.Wt= 63 g/mole.