1. MATTER CHAPTER 1

4. EXERCISE: Give the concentration of a AgNO 3 solution with 5.00 g of solute dissolved in 250.0 mL solution. [Ans: 20.0 gL -1 ]

8. Exercise: 1. Dextromethorphan, a cough mixture, contains 3.0 % w/v of Dextromethorphan hydrobromide with 0.2 % of methylparaben as one of the preservatives What is the mass in gram of Dextromethorphan hydrobromide in 90 mL of the cough mixture? ANS: 2.7 g

10. EXERCISE: 1. A KOH solution containing 7.00 g KOH (RAM KOH = 56.0)dissolved in 500.0g water to form a 502.0 mL solution. Find its molality. Ans: 0.25 m

14. EXERCISE: 1. The concentration of calcium ions in blood is 100.0 ppm. Calculate the mass of calcium ions in 500.0 g of blood. Ans: 0.05 g

15. 2. What is the volume of 1.55 ppm solution formed by dissolving 454 g of NaF in water? Ans: 2.93 x 10 5 L C ppm = mass of solute (mg ) volume of solution (L) 1.55 (mg/L) = 454 g x 10 3 mg/g volume of solution (L) volume of solution = 454 x 10 3 mg 1.55 mg/L = 2.93 x 10 5 L

16. Exercise: 1.Calculate the mass of NaCl used to make a salt 1.00 ppm solution of 500 mL ? Ans: 5.00 x 10 -4 g Ans: mass of solute (mg ) = 1.00 mg L -1 0.500 L mass of NaCl = 0.500 L x 1.00 mg L -1 = 0.500 mg = 0.500 mg x 10 -3 g/mg = 5.00 x 10 -4 g

21. Dilution

25. 1. A solution contains 55 g toluene, C 7 H 8 and 55 g of bromobenzene,C 6 H 5 Br. Find the mole fraction of each component. Ans: 0.63; 0.37 2. The concentration of Cl - ion in a water source is about 0.0 1 g per 1000g of water. Find the concentration of Cl - ion in the water in parts per million, ppm. Ans: 10.0 ppm 3. Express the concentration of a 0.10 M HCl solution in g/L .[H = 1.00; Cl = 35.5 ] [Ans:3.65 g/L]

27. 7 . How many moles of NaOH are required to prepare a solution of 250 mL with 0.010M ? How many grams of NaOH are required ? If the solvent used is 249 mL , find its molality

33. Review of last lecture 1.Concentration ( g/L ) 2. Molarity ( Mol/L @ M ) 3.Molarity ( Mol/L @ M ) 4. % by mass(w/w) 5. % w/v ( weight per volume )

42. SOLUTION: a. Cl 2 + C 2 O 4 2-  Cl - + CO 2 Step 1 : Cl 2 + C 2 O 4 2-  Cl - + CO 2 ON 0 +3 -2 -1 +4 -2 Step 1 +2 : Cl 2  2 Cl - [R] C 2 O 4 2-  2 CO 2 [O] Step 3 + 4 : Cl 2 -  2 Cl - C 2 O 4 2-  2 CO 2

43. Step 5 : Cl 2 +2e  2 Cl - --------- I) C 2 O 4 2-  2 CO 2 +2e--------- ii) Step 6: I) + ii) Cl 2 + C 2 O 4 2-  2Cl - + 2CO 2 b)SO 3 2- + MnO 2 + H +  SO 4 2- + Mn 2+ + H 2 O Steps: 1 S O 3 2- + Mn O 2  S O 4 2- + Mn 2+ +4 +4 +6 +2

44. Step 1&2 SO 3 2-  SO 4 2- [O] MnO 2  Mn 2+ [R] Step 3 SO 3 2- + H 2 O  SO 4 2- [O] MnO 2  Mn 2+ + 2 H 2 O [R] Step 4 SO 3 2- + H 2 O  SO 4 2- + 2H + [O] MnO 2 + 4H +  Mn 2+ +2 H 2 O [R] STEP 5 SO 3 2- + H 2 O  SO 4 2- + 2H + + 2e [O] ---I) MnO 2 + 4H + + 2e  Mn 2+ +2 H 2 O [R]– ii) Step 6 i) + ii) SO 3 2- + MnO 2 + 2H +  SO 4 2- + Mn 2+ + H 2 O

57. iii) Find the mole of ZnS formed and hence deterrmine its mass. iv) Find the mole of excess reactants left over after the reaction is complete. v) From iv) determine the mass of excess reactant left over.

58. a) Mole of Zn = 6.35 g / 65.4 g mol -1 =0.0971mol mole of S = 8.0 g / 32.0 g mol -1 = 0.25 mol b) To find limiting reagent: Zn S ratio of n / the stoi.coefficient : 0.0971/1 : 0.25/1 = 0.0971 : = 0.25 The limiting reactant is Zn with the smaller ratio The excess reactant is S Solution

59. c) From equation, 1 mol Zn ≡ 1 mol of ZnS 0.0971 mol Zn ≡ 0.0971 mol ZnS = 0.0971 mol x [65.4 + 32.0] g mol-1 = 9.46 g d) 1 mol of Zn (limiting react ant) ≡ 1mol S (excess reactant) 0.0971 mol Zn ≡ 0.0971 mol S e) Excess S left over after the reaction = 0.25 – 0.0971 = 0.15 3mol S mass of excess S left over = 0.153 mol x 32.0 g mol-1 = 4.89 g

63. a) Mole of CaCO 3 = 100g /100.0 g mol -1 = 1.00 mol mole of HCl = 2.0 M x 0.400L = 0.80 mol b) To find limiting reactant: CaCO 3 HCl ratio of n / the stoi.coefficient : 1.00/1 : 0.80/2 = 1.00 : = 0.40 The limiting reactant is HCl with the smaller ratio The excess reactant is CaCO 3 Solution

64. d) 2 mol of HCl (limiting react ant) ≡ 1mol CO 2 0.80 mol HCl ≡ 0.80/2 mol CO 2 = 0.40 mol Volume of CO 2 released at STP = 0.40 mol x 22.4 L mol -1 =8.96 L e) Excess CaCO 3 left over after the reaction = 1.00 - 0.40 = 0.60 mol mass of excess CaCO 3 left over = 0.60 mol x 100.0 g mol-1 = 60 g

81. Solution i. 5Fe 2+ (aq) + MnO 4 - + 8 H + 5Fe 3+ (aq) + Mn 2+ + 4H 2 O Method 2 Mole KMnO 4 @ MnO 4 - = 0.02 x 0.09295 = 0.001859 1 mole MnO 4 - requires 5 moles Fe 2+ Mole Fe 2+ or FeSO 4 = 0.001859 x 5 = 0.009295 Mass of Fe in iron ore = 0.09295 x 56 = 0.521 g % Fe in sample = 33.56 %