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online practice problems - 44 CALCULATIONS INVOLVING...

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44 One litre = 1 000 mL. Since 1.0 mL = 1.0 cm , then 1.000 litre = 1000 cm . 13 3 One decimetre (dm) = metre = 10 cm. (1.0 dm) = (10 cm) . 33 1.0 dm = 1000 cm = 1.000 L. 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 ), or as Molarity (M), which means the same. In practical terms, however, -1 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 dm , or 3 cubic decimetres ). Millilitres (mL) are also used widely, but may have to be converted to 1 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 ) = or C = -1 This can also be written: Number of moles of solute = Concentration (mol. L ) x Volume of solution (litres) -1 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.
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45 Example one: A solution of sodium hydroxide has a concentration = 0.50 mol. L (also written as -1 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? 32 Lead nitrate: formula = Pb(NO ) , 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?
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This note was uploaded on 03/07/2012 for the course CHEM 1201 taught by Professor Cook during the Spring '08 term at LSU.

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online practice problems - 44 CALCULATIONS INVOLVING...

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