concentration of solution

1. Concentration of solution based on volume
and mass
units, Calculations of ppm, ppb and dilution of
solutions
By : Pranav Ganesh Dalvi

2. Different ways of Expressing Concentration
of Solutions
Sample Footer Text 20XX 2

3. A. volume basis
◦ a. Molarity and Normality
◦ It is often helpful to know how many moles of solute are present in one liter of solution especially when this
solutions are involved in chemical reactions. Molarity and Normality describes the number (moles) off
reactants or products dissolved in one liter of solution
◦ Molarity
◦ Molarity : M= moles of solute contained in one litre of solution
◦ The molecular weight of glucose sugar is 180 g/mole . If 360 gram of glucose he’s dissolved in enough
water to make one litre of solution , the concentration of glucose is 2.00M

4. Numericals
◦ Q1. 2.4 gram of sodium carbonate dissolved in 245ml of solution calculate molarity of solution(0.09241M)
◦ Q2 . Prepare 700ml of 0.25 molar solution of sodium hydroxide. calculate weight of NaOH solution in
gram (7g)
◦ Q3. 10 gram of NaOH with volume of 0.2 dm3 . calculate molarity .(1.25M)
◦ Q4 . 100ml of LiCl contains 0.637 gram of LiCl (molecular weight 42.44) density is 1.003 g/ml.( calculate
molarity(0.15009M)
◦ Q5. Concentrated HNO3 specific gravity 1.42 contains 69% (w/w) . Calculate molarity(10.951M)

5. A. volume basis
◦ Normality
◦ Normality : N= moles of reactive units per litre (equivalents per litre)
◦ Also defined as number of equivalent weight of solute in one litre of the solution
◦ Where molarity describes the mole of a complete substance per litre of solution normality describes only the
mole of reactive species per litre of the solution
◦ Normality is always a multiple of molarity it describes the “equivalent” moles of reactant involved in
chemical reaction

6. Normality
◦ Equivalent weight : the weight of the substance which provides the Avogadro number of reacting unit. The
type of reacting unit depends on the reaction involved
◦ The type of reaction involved are:
1. Neutralization reaction
2. Precipitation and complexometric reactions
3. Redox reaction
◦ Normality in acid base reaction:
◦ In an acid base reaction, normality is a major of the protons (H+) or hydroxides(OH-) that react with one
another. Equivalent weight of an acid is that weight of it which contain 1 gram atom of replaceable
hydrogen. Equivalent weight of monobasic acid is equal to its molecular weight

7. Normality
◦ Normality of redox reagents:
◦ In reduction oxidation reactions, electrons move from oxidized atom to reduced atoms. To calculate the
normality of redox reactant the number of electrons donated or accepted per mole of the reactant. The mass
of a reactant that donates or accepts one mole of electrons during the reaction is the “equivalent weight” of
this reactant.
◦ Dividing the total grams of a reactant by the equivalent weight yields the “equivalents” of this reactant
consider the following redox reaction:
◦ Zn + 2Cu2+ Zn2+ + 2Cu+
◦ Observation
1. Each zinc atom donates two electron during the reaction, while each atom of copper accepts only one
electron.
2. ½ mole of zinc donates one mole of electron so the equivalent weight of zinc is ½ of the atomic weight
:65.4 / 2 which is equal to 32.6 gram per equivalent

8. Normality
◦ Normality of redox reagents:
3. However, since 1 mole of copper atom accepts only one mole of electrons the equivalent weight of copper
in this reaction equals its atomic weight: 63.5 / 1= 63.5 gram per equivalent

9. Normality
◦ Precipitation and complexometric titrations
◦ Related to charge carried by the ions
◦ The basic advantage in dealing with the equivalent is any two species will always have the stoichiometric ratio
1:1 which may not be possible in the case of molar ratio
◦ valency is considered here
◦ VANA=VBNB

10. Numericals
1. 3.6 gram of oxalic acid(H2C2O4⋅2H2O) are dissolved and volume made-up to 150 ml. calculate the
molarity of the solution, normality of the solution on basis of acid and base titration, normality on the
basis of redox titration.(0.38N)
2. 1.5 dm3 of 0.12 N HCL mixed with 0.2 dm3 of 1.2 M calculate g/ dm3 of HCL before and after dilution
3. 7.03 grams of dibasic acid of molecular weight 98 are dissolved in 0.8 litre of the solution calculate the molarity and
normality of the acid solutions(0.1793N ,0.08967M)
4. Calculate equivalent weight of a. oxalic acid with NaOH(63,40) b. oxalic acid with KmnO4(63,31.3068) c. Kmno4 in
acidic medium(31.3068) d. AgNO3 reaction with NaCl(169.87,58.44)

11. Formality
◦ The solution which contain one formula weight of this salute in one litre of the solution
◦ F= number of formula weight of solute/ volume of solution in litre
◦ Number of formula weight of solute= amount in grams of the solute/ formula weight of the solute
◦ F=W/FxV
◦ Difference between formula mass and molar mass is that the formula mass of amolecule or a compound is
the sum of the atomic weights of the atoms in the empirical formula while molar mass is the mass in grams
of 1 mol of substance

12. B. Weight basis
◦ Molality
◦ Molality is defines as number of moles of solute dissolved in mass of solvent in 1 kg
◦ m= amount in grams of the solute/ molar mass of the solute x weight of solvent in kg
◦ m=WB/MB X WA
◦ Mole fraction :
◦ Defined as ratio of number of moles of that component to the total number of moles present in the system
◦ XA=nA / nA+nB
◦ XB=nB / nA+nB

13. Numericals
1. 150 g of Na2SO4.10H2O are dissolved in 100 ml of the solution .calculate formality of the solution and
molarity of Na+ and SO4
2- ions
2. Calculate formality of solutions prepared by dissolving 1.96 x 10-3 kg of sulphuric acid in water to form 1
L of the solution
3. 5 x10-2 kg of A(M.W=0.2kg) and 5 x10-2 kg of B are mixed together . The mole fraction of A in the
mixture is 0.7410 .calculate molecular weight of B

14. Parts per million(ppm) and Parts per
billion(ppb) and numericals

15. Dilution of solution

16. Numericals

17. More Numericals

18. Thank you