Dr. Sukhdeep Singh
SOLUTIONS, CONCENTRATIONS, STOICHIOMETRY
Aim:




Knowledge about (i) % Mass percentage or volume percentage (ii) N
normality (iii) M morality (iv) m molality (v) ppt vs ppm.
Raoult’s law for solutions containing volatile and non-volatile solutes.
Ideal and non-ideal solutions, meaning of colligative property, elevation of
boiling point, depression in freezing point.
Introduction to osmosis, osmotic pressure and diffusion and azetropic mixture.
Numerical Problems:
1) What weight of ferrous ammonium sulphate [(NH4)2Fe(SO4)2·6H2O)] is needed to
prepare 100mL of 0.1 normal solution?
2) How many grams of H2SO4 are required to make a decinormal acidic solution.
3) If 2.2 g of oxalic acid (H₂C₂O₄) are dissolved in 500 mL of solution with density 1.1
g/mL. (i) What is the mass percent of oxalic acid in solution? (ii) Considering the
complete ionization of oxalic acid what is the Normality of the solution?
4) An experiment requires addition of 0.184 NaOH in aqueous solution. How many
milliliters of 0.15 M NaOH should be added to the reaction vessel?
5) Calculate the molarity and molality of a 15% solution of H2SO4 (density of H2SO4
solution = 1.1 g cm-3).
6) You are given 5 mL of 14.8 M NH3. What will be the final volume after this solution
is diluted with water to get 1 M NH3 solution.
7) 1000 g of aq. solution of CaCO3, contains 10 g of carbonate. What is the
concentration in ppm?
8) A 500 g tooth paste sample has 0.2 g fluoride concentration. What is the conc of F- in
terms of ppm level?
Answers:
1)
2)
3)
3.92 g
4.9 g
Homework Part 1
Density of solution = 1.1 g/L
Volume of solution = 500 mL
Mass of solution = Density x volume
= 1.1 g/mL x 500 mL = 550 g
Mass of oxalic acid = 2.2 g/ 550g x 100 = 0.4 %
Part 2
Dr. Sukhdeep Singh
4)
5)
6)
7)
8)
Normality = a/ Eq.mass X 1000/V
2.2/45 X 1000/ 500 = 0.098
Homework Moles of solute required = 0.184g / 40 g mol-1 = 0.0046 mol
Volume of solution = Moles of solute/ Molarity
= 0.0046 mol/ 0.15 mol L-1 = 0.0307 L or 30.7 mL
Molality
Moles of H2SO4 = 15 g/ 98 g mol-1 = 0.153 mol
Mass of the solvent 85g or 0.085 Kg
molality = 0.153 mol/0.085 Kg = 1.8 m
Molarity
Volume of solution = 100 g / 1.10 g cm -3 = 90.91 cm 3 or 0.0909 L
Molarity = 0.153 mol/ 0.0909 L = 1.68 M
M1V1 = M2 V2
14.8 M x 5 mL / 1.00 M = 74.0 mL
10000 ppm
400 ppm