Concentration Calculations and Equivalent Mass in Chemistry

2025-2024

Session Objectives 1. Equivalent mass 2. Normality 3. Molarity 4. Percentage concentration

Equivalent Mass Acid Equivalent mass Base Salt

Equivalent Mass of Acid Equivalent mass of acid = Molecular mass of acid Number of replacable H (Basicity) + Example: Equivalent mass of HCl and H2SO4 HCl + H + Cl + H SO 2H + SO 2 4 4 1 + 35.5 1 2 Equivalent mass of HCl = = 36.5 1 + 32 + 4 16 Equivalent mass of H SO = = 49 2 4 2

Equivalent Mass of Base Equivalent mass of base = Molecular mass Number of replacable OH (Acidity) Example: Equivalent mass of NaOH and Ca(OH)2 NaOH Na + + OH ++ Ca(OH) Ca + 2OH 2 23 + 16 1 + + 1 Equivalent mass of NaOH = = 40 40 2 16 2 + 2 1 Equivalent mass of Ca(OH) = = 37 2

Equivalent mass of salt Equivalent mass of salt = Molecular mass Total number of positive or negative charge Example: Equivalent mass of NaCl and MgCl2 NaCl + Na + Cl 23 + 35.5 1 Equivalent mass of NaCl = = 58.5 ++ MgCl Mg + 2Cl 2 240 + 2 2 35.5 Equivalent mass of MgCl = = 47.5 2

Concentration of solutions (1) Normality Number of equivalents of solute present in one litre of solution. Equivalent of solute N Volume of solution in litre = Mass of solute 1000 volume (in ml) N = Equivalent mass of solute Equivalents V(in litre) Also N = Equivalents = N x V (in litre) Milli equivalents = N x V (in ml)

Examples Q1. Find the normality of H2SO4 having 49g of H2SO4 present in 500 ml of solution. Solution: Mass of solute Equivalent mass 1000 N = volume (in ml) 49 98 2 1000 N = = 2N 500

Most important point about equivalents Equivalent and milliequivalents of reactants reacts in equal number to give same number of equivalents or milliequivalents of products separately. Example: 2NaOH + H SO Na SO + 2H O 2 4 2 4 2 2 Equivalents 2 Equivalents 2 Equivalents 2 moles 1 mole 1 mole 2 mole

Q2. 20 ml of 0.1 N BaCl2 is mixed with 30 ml of 0.2 N Al2(SO4)3. How many gram of BaSO4 are formed? BaCl + Al (SO ) AlCl + BaSO 2 2 4 3 3 4 Solution: By equivalent method, no need of balancing the equation. Because equivalents of reactants and products are same. BaCl Al (SO ) + AlCl + BaSO 2 2 4 3 3 4 20 1000 0.1 Equivalents of BaCl2 = = 2 x 10 3 30 1000 0.2 Equivalents of Al2(SO4)3 = = 6 x 10 3

Solution contd- Since equivalents of Al2(SO4)3 is in excess, hence equivalents of BaSO4= equivalents of BaCl2 = equivalents of AlCl3 = 2 x 10 3 mass of BaSO4 = Equivalents x equivalent mass 233 2 3 = 2 10 = .233g If we will discuss this problem through mole concept, then we have to balanced the equation.

Molarity Number of moles of solute present in one litre of solution. Moles of solute Volume (in litre) M = Moles of solute Molecular mass 1000 M = volume (in ml) Moles = Molarity x volume (in litre) Milli moles = Molarity x volume (in ml)

Q3. Calculate the molarity of a solution of NaOH in which 0.40g NaOH dissolved in 500 ml solution. Solution: 0.40 40 M = 1000 500 = 0.02 M

Relation between normality and molarity Mass of solute Molecular mass n factor 1000 N = volume (in ml) N = M x n factor For HCl, n = 1 H2SO4, n = 2 H3PO4, n = 3 NaOH, n = 1 Ca(OH)2, n = 2 For monovalent compound (n = 1) Normality and molarity is same.

Illustrative Problem Calculate molarity of 0.6 N AlCl3 solution. Solution: +++ AlCl Al + 3Cl 3 n = 3 0.6 3 M = = 0.2M

Q4. 0.115 g of pure sodium metal was dissolved in 500 ml distilled water. The molarity of the solution would be : (Na = 23) (a) 0.01 M (b) 0.00115 M (c) 0.023 M (d) 0.046 M Solution: Mass of solute Molecular mass of solute 1000 M = Volume in ml 0.115 23 = 1000 = 0.01M 500 Hence, answer is (a)

Q5. Calculate the molarity and normality of a solution containing 0.5 g of NaOH dissolved in 500 ml. Solution: Mass of solute Molecular mass 1000 Molarity = ( ) Volume in ml 0.5 40 1000 500 Mass of solute Equivalent mass 0.5 40 500 1 = = 0.025 M 1000 Normality N = ( ) Volume in ml = 1000 = 0.025 N Or for monovalent compound like NaOH normality and molarity are same.

Q6. Calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass. Solution: 95% of ethanol by mass means 95 g ethanol present in 100 g of solution. Hence, mass of water = 100 95 = 5 g 95 46= 2.07 moles Moles of C2H5OH = 5 18= 0.28mol Moles of water(H2O)=

Solution 0.28 + = 0.88 Mole fraction of water = 0.28 2.07 Mole fraction of ethanol = 1 0.88 = 0.12

Q7. A solution contains 25% of water, 25% of ethanol and 50% of acetic acid by mass. Calculate the mole fraction of each component. Solution: 25x + 25x + 50x = 100 x = 1 Mass of water = 25 g Mass of ethanol = 25 g Mass of acetic acid = 50 g Moles of water =25 = 1.388 moles 18

Solution Moles of ethanol =25 = 0.543 moles 46 Moles of acetic acid =50=0.833 moles 60 1.388 2.764= Mole fraction of water = 0.502 Mole fraction of ethanol= 0.543/2.764= 0.196 Mole fraction of acetic acid = 1 0.503 0.196 = 0.301

Q8. 20 ml of 10 N HCl are diluted with distilled water to form one litre of the solution. What is the normality of the diluted solution? Solution: N1V1 = N2V2 20 1000 1000 10 = N 1000 2 N2 = 0.2 N

Q9. How many milliliters of distilled water must be added to 985 mL of 0.01295 N sodium thiosulfate to get a solution with a concentration of 0.0125 N sodium thiosulfate? Na2S2O3 N1 x V1 = N2 x V2 0.01295 N X 985 mL = 0.0125 N X V2 0.01295 x 985 0.0125 1020.46 mL = V2 - 985.0 mL 35.46 mL = V2 Water to be Added

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