1. A. Solid in liquid solubility e.g., NaCl or Sugar in water Concentration Expressions of Solutions Molarity(M): Moles of solute in 1000 ml of solution Molality* (m): Moles of solute in 1000 gm of solvent Mole Fraction(  A ): Ratio of the moles of one constituent (solute or solvent) of a solution to the total moles of all constituents (solute and solvent). Mole percent: is obtained by multiplying mole fraction by 100. Number of Moles: Weight in grams divided by molecular weight in gm / mole * Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution.

2. % (w/w) = % (w/v) = % (v/v) = Other concentration expression: % Concentration

3. Example: An aqueous solution of ferrous sulphate (FeSO 4 ) was prepared by adding 41.5 gm of FeSO 4 to enough water to make 1 L of solution at 18 o C. The density of the solution is 1.0375 gm/mL, and the molecular weight of FeSO 4 is 151.9 gm/mol and molecular weight of H 2 O is 18.02 gm/mol. Calculate A) the Molarity of FeSO 4 ; Number of moles of FeSO 4 = Weight ÷ molecular weight Number of moles of FeSO 4 = 41.5gm ÷ 151.9 gm/mol = 0.2732 moles Molarity (M) = No. of moles in liters of solution Molarity = 0.2732 ÷ 1 L = 0.2732 M B) the Molality of FeSO 4 ; Molality = No. of moles in 1000 gm of solvent No. of moles = 0.2732 moles Volume of solution= 1000 ml, But weight of solvent is not known. Weight = volume X density Weight of solution = 1000 ml X 1.0375 gm/ml = 1037.5 gm Weight of solvent = 1037.5 gm- 41.5 gm = 996 gm=0.996 kg Molality = 0.2732 ÷ 0.996 kg= 0.2743 m

4. Example-continued: An aqueous solution of ferrous sulphate (FeSO 4 ) was prepared by adding 41.5 gm of FeSO 4 to enough water to make 1 L of solution at 18 o C. The density of the solution is 1.0375 gm/mL, and the molecular weight of FeSO 4 is 151.9 gm/mol and molecular weight of H 2 O is 18.02 gm/mol. C) the mole fraction and mole percent of water. Mole fraction (X 1 ) of water = moles of water/ (moles of water +moles of FeSO 4 ) Moles of water = 996 gm ÷ 18.02 gm/mol = 55.27 moles Mole fraction (X 1 ) of water = 55.27 / (55.27 + 0.2732 )= 0.9951 Mole percent of water = mole fraction X 100 Mole percent of water = 0.9951 X 100 = 99.51% D) the mole fraction and mole percent of FeSO 4 ; Mole fraction (X 2 ) of FeSO 4 = moles of FeSO 4 / (moles of water +moles of FeSO 4 ) Mole fraction (X 2 ) of FeSO 4 = 0.2732 / (55.27 + 0.2732 ) = 0.0049 mole percent of FeSO 4 = mole fraction X 100 mole percent of FeSO 4 = 0.0049 X 100 = 0.49% N.B Sum of X 1 and X 2 = 0.9951+0.0049 = 1.00

5. Example-continued: An aqueous solution of ferrous sulphate (FeSO 4 ) was prepared by adding 41.5 gm of FeSO 4 to enough water to make 1 L of solution at 18 o C. The density of the solution is 1.0375 gm/mL, and the molecular weight of FeSO 4 is 151.9 gm/mol and molecular weight of H 2 O is 18.02 gm/mol. E) the percent w/w of FeSO 4. the percent w/w of FeSO 4 = (weight of FeSO 4 ÷ weight of solution)X100 the percent w/w of FeSO 4 = (41.5 gm ÷ 1037.5)X100 = 4% w/w F) the percent w/v of FeSO 4. the percent w/v of FeSO 4 = (weight of FeSO 4 ÷ volume of solution)X100 the percent w/v of FeSO 4 = (41.5 gm ÷ 1000 ml)X100 = 4.15% w/v

6. Other descriptive terms of solubility: Very solubleone gm of solute / one ml of solvent Freely solubleone gm of solute/ from < 1 to 10 mls of solvent solubleone gm of solute/ from <10 to 30 mls of solvent sparingly soluble one gm of solute/ form < 30 to 100 mls of solvent slightly soluble one gm of solute/ form < 100 to 1000 mls of solvent very slightly soluble one gm of solute/ form < 1000 to 10000 mls of solvent practicaly insoluble or insoluble one gm of solute/ more than 10,000 mls of solvent