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Tutorial 1 Answer Sheet
1 - Calculate the following: a. The weight in grams of 0.45 moles of glucose (C 6 H 12 O 6 ) No. of moles = wtg /MWT thus, wtg = No. of moles × MWT MWT of glucose = (12×6) + (1×12) +(16×6) = 180g/mole wtg = 0.45 ×180 = 81 g b. The weight in grams of 1 × 10 23 molecules of NaCl. 1 mole has 6.023 × 10 23 ? mole has 1 × 10 23 = 0.166 mole MWT of NaCl = (1× 23) + (1× 35.5)= 58.5g/mole wtg = 0.166 × 58.5 = 9.71 g c. The number of molecules in 2.25 g of glycine. MWT of glycine = (2×12) + (1×14) + (2×16) + (5×1) = 75g/mole. 1 mole has 75g ? mole has 2. = 0.03 mole Since 1 mole has 6.023 × 10 23 0.03 mole has? molecules =0.03 × 6.023 × 10 23 = 0.18 × 10 23 2 - Calculate the normality of the following solutions: a. 250 ml of HCl containing 18.25 g of HCl. N = No. of equivalents / V(L) No. of equivalents = wtg of solute / equivalents weight EW= MWT of solute / n MWT of HCl= (1×35.5) + (1×1) = 36.5g/mole n = 1 EW= MWT of solute / n = 36.5 / 1 = 36. No. of equivalents = wtg of solute / equivalents weight = 18.25 /36. = 0.
N = No. of equivalents / V(L) = 0.5 / 0. =2 normal b. 49 g of H 2 SO 4 in 250 ml. MWT of H 2 SO 4 = (2×1) + (1×32) + (4×16) = 98g/mole n= EW = MWT of solute / n = 98/ 2 = 49 No. of equivalents = wtg of solute / equivalents weight = 49 / = 1 N = No. of equivalents / V(L) = 1 / 0. = 4 normal 3 - 12.25 g of phosphoric acid was dissolved in water and the volume made up to 100 ml, calculate: a. The normality of the solution. b. The molarity of the solution. N = M × n M = No. of moles of solute / V(L) No. of moles = wtg /MWT MWT of H 3 PO 4 = (3×1) + (1×31) + (4×16) = 98g/mole. No. of moles = wtg /MWT = 12.25 / 98 = 0.125 mole M = No. of moles of solute / V(L = 0.125 / 0. = 1.25 molar n = 3 N = M × n = 1.25 × 3 = 3.75 normal 4 - 20 g of NaCl is dissolved in 200 ml water, what is its W/V%? 20 g in 200 ml ? in 100 ml
N= 1 × 3
= 3 Osmolarity 8 - How would you prepare 0.2 L of 0.3 MgCl 2 W/V% solution. 0.3 g in 100 ml is 0.3% W/V% but since 200 ml is needed ? g in 200 ml = (0.3 × 200) / 100 = 0.6 g of MgCl 2 0.6 g of MgCl 2 is dissolved in a little volume of distilled water then the volume was made up to 200ml with distilled water 9 - A solution was prepared by dissolving 8 g of solid ammonium sulfate (MWT = 132.14) in 39.52 ml of water. Express the concentration in terms of: g/L, M, N, W/V%, mg%, osmolarity. g/L 8g --------> 39.52 ml ? ---------> 1000 ml ? = (8 × 1000)/39.52 = 202.4 g/l M = no. of moles/volume(L) no. of moles = wt/MWT = 8/132.14 = 0.0605 mole volume(L) = 39.52 ml/1000 = 0.03952 L M= 0.0605/0. = 1.53 molar N = n × M = 1 × 1. = 1.53 N W/V% = wt in g/ volume in 100 ml of solution 8g --------> 39.52 ml ? ---------> 100 ml ? = (8 × 100)/39.52 = 20.24 % mg% from W/V% = 20.24% mg% =20.24 × 1000 = 2024 mg% Osmolarity = n × M
= 2 × 1.
= 3.06 osmolar 10 - A solution contains 15 g of CaCl 2 in a total volume of 1 9 0 ml. Express the concentration of this solution in terms of: g/L, M, W/V%, mg%, osmolarity. g/L 15g -------- > 190 ml ? --------- > 1000 ml ? = (15 × 1000)/190 = 78.9 g/l M = no. of moles/volume(L) no. of moles = wt/MWT = 15/(40 + (35 × 5)) = 0.135 mole volume(L) = 190 ml/1000 = 0.19 L M= 0.135/0. = 0.711 molar W/V% = wt in g/ volume in 100 ml of solution 78.9g -------- > 1000 ml ? --------- > 100 ml ? = (78.9 × 100)/1000 = 7.89 % mg% from W/V% = 7.89% mg% =7.89 × 1000 = 7890 mg% Osmolarity = n × M = 3 × 0. = 2.134 osmolar
Tutorial 2 Answer Sheet
- How many ml of 0.8 M acetic acid (CH 3 COOH) are needed to prepare 200ml of 0.4N acetic acid? N = M × n
= (45.08 × 100) / 92
= 49g of solution V= wt / ρ = 49/ 1. = 26.6ml So 26.6ml of the stock solution is taken then complete up the volume to 2 liters with distilled water.
- Calculate the molarity of H 2 SO 4 which has a molality of 6.8 molal, P = 1.48 g/ml? Molality means 6.8 mole of solute in 1000 g of solvent. MWT of H 2 SO 4 = (2×1) + (1×32) + (4×16) = 98g/mole. No. of moles = wtg /MWT thus wtg = No. of moles × MWT = 6.8 × 98 = 666.4g The weight of solution = weight of solvent + weight of solute. = 1000 + 666. = 1666.4 g V= wt / ρ = 1666.4 / 1. = 1125.9 ml Since 6.8 mole of solute in 1125.9 ml of solution ? mole of solute in 1000 ml of solution = 6.04 molar
- A solution of H 2 SO 4 is 4% W/W%, density is 1.84g/ml. Calculate the molarity, normality and molality? Molarity MWT of H 2 SO 4 = (2×1) + (1×32) + (4×16) = 98g/mole. Since 4% W/W is 4 g H 2 SO 4 in 100g solution. No. of moles = wtg /MWT = 4 / 98 = 0.04 mole V= wt / ρ = 100/ 1. = 54.35 ml of solution So 0.04 mole in 54.35ml solution
? Mole in 1000 ml of solution = (0.04 × 1000) / 54. = 0.74 molar Normality N = M × n n= 2 N= 0.74 × 2 = 1.48 normal Molality Since the weight of solution = weight of solvent + weight of solute. Thus, the weight of solvent = weight of solution - weight of solute. = 100g – 4 g = 96g 0.04 mole of solute in 96 g of solvent ? mole of solute in 1000 g of solvent No. of moles of solute 1000 g of solvent = ( 0.04 × 1000) / 96 = 0.42 moles The molality is 0.
- Describe how to prepare a 400 ml ,1:8 dilution of a disinfectant solution from a stock solution provided using water as your diluent. Since DF=Vf / Vi thus Vi = Vf / DF DF is 8 Vi = 400 / 8 = 50 ml. To prepare, we need 50ml of the stock solution and make up the volume to 400ml with water
- You are provided with 3ml of a 100 mg/ml stock solution of ampicillin and requested to prepare dilute it a final concentration of 25mg/ml and final volume of 200μl. Calculate the volume of stock solution needed? Describe the preparation process. C 1 × V 1 = C 2 × V 2 C 1 = 100 mg/ml, C 2 = 25mg/ml, V 2 = 200μl, V 1 =? 100 × V 1 = 25 × 200
For tube 7 = 10 × 10 × 10 × 10 × 10 × 10 × 10 = 10000000. The concentration is 0.4/10000000 = 4 × 10
