• Document: Number of moles of solute = Concentration (mol. L ) x Volume of solution (litres) or n = C x V
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44 CALCULATIONS INVOLVING SOLUTIONS INTRODUCTION AND DEFINITIONS Many chemical reactions take place in aqueous (water) solution. Quantities of such solutions are measured as volumes, while the amounts of solute present in a given volume of solution are the concentrations of the solutions. The concentrations of solutions are commonly stated in moles (of solute) per litre (of solution)(mol. L-1), or as Molarity (M), which means the same. In practical terms, however, amounts of solutes have to be measured as mass, for example in grams. It is not possible to measure the number of moles of a solid directly with a balance or any other instrument. Concentrations of solutions can also be expressed, therefore, in grams (of solute) per litre (of solution) (g. L-1). Grams per litre is therefore a practical way to express concentration of solute in a solution. Moles per litre, or molarity, is more important in making calculations and predictions about chemical reactions. Volumes of solutions are defined in litres, (L), (which is sometimes expressed as dm3, or cubic decimetres1). Millilitres (mL) are also used widely, but may have to be converted to litres for some calculations. FORMULAE USEFUL FOR CALCULATIONS: It is important, therefore, to be able to convert grams to moles and moles to grams for any substance. Number of moles = or n = This can also be written: Actual mass (grams) = Number of moles x Molar mass, or m = n x M Concentration of a solution can be expressed as a formula: Concentration (in mol. L-1) = or C = This can also be written: Number of moles of solute = Concentration (mol. L-1) x Volume of solution (litres) or n = C x V It is suggested strongly that these formulae be remembered as word formulae, rather than as algebraic formulae. With word formulae, the student is remembering the meanings of the parts of each formula. It is easy, with the algebraic formulae, to confuse n and m and M. 1 One litre = 1 000 mL. Since 1.0 mL = 1.0 cm3, then 1.000 litre = 1000 cm3. One decimetre (dm) = metre = 10 cm. (1.0 dm)3 = (10 cm)3. 1.0 dm3 = 1000 cm3 = 1.000 L. 45 Example one: A solution of sodium hydroxide has a concentration = 0.50 mol. L-1 (also written as 0.50M). How many grams of sodium hydroxide are dissolved in 1.000 L of solution? Calculate first the number of moles in the 1.000 L of solution. Concentration = 0.50M = = Number of moles of solute = 0.50 x 1.000 = 0.50 mol Sodium hydroxide: formula = NaOH, molar mass = (23.0 + 16.0 + 1.0) g = 40.0 g One mole of NaOH = 40.0 g, so 0.50 mol of NaOH = 20.0 g. 20.0 g of solid sodium hydroxide dissolved in 1.000 L solution makes a solution of concentration = 0.50M. Example two: What mass of lead nitrate should be dissolved in 250 mL of water to make a solution of concentration 0.040M? Lead nitrate: formula = Pb(NO3)2, molar mass = 331.2 g 250 mL = 0.250 L Concentration = 0.040M = = Number of moles of solute = 0.040 x 0.250 = 0.010 mol Mass of lead nitrate required = (0.010 x 331.2) g = 3.31 g. Example three: What is the concentration of the solution if 10.0 g of sodium carbonate is dissolved in 200 mL of solution? Sodium carbonate: formula = Na2CO3, molar mass = 106.0 g 200 mL = 0.200 L Number of moles of Na2CO3 in 10.0 g = = 0.0943 mol. Concentration = = = 0.47 M Example four: What volume of solution must be made if 5.10 g of silver nitrate is to be dissolved to make a solution of concentration = 0.050M? Silver nitrate: formula = AgNO3, molar mass = 169.9 g Number of moles of silver nitrate = = = 0.0300 mol Concentration = 0.050 = Volume = L = 0.600 L = 600 mL Calculations of this kind should be practised until they can be done quickly and easily. 46 Exercises: 1. How many moles of calcium hydroxide are needed to make 2.000 L of 0.001M solution? What mass of calcium hydroxide would be required? 2, What volume of 0.05M potassium sulfate contains 50.0 g of potassium sulfate? (Hint: first convert the number of grams to number of moles.) 3. What is the concentration of magnesium sulfate if 20.0 g of crystals (MgSO4.7H2O) are dissolved in 500 mL of solution? 4. What is the mass of hydrogen chloride in 2.50L of 10.5M hydrochloric acid solution? (Hydrochloric acid is a solution of hydrogen chloride gas in water. "Concentrated hydrochloric

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