IB Topic 1: Quantitative Chemistry 1.5 Solutions  Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 )  Solve problems involving concentration, amount of solute and volume of solution.

Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 ) Solution: Homogeneous mixtures of two or more substances. Most common is solid, liquid or gas dissolved in a liquid (usually water). These are called aqueous solutions. Can have other solutions such as solid-solid (alloy) or gas-gas (air). Solute: The dissolved particles. Usually the substance in the least amount. Solvent: The dissolving medium. Usually the substance in the greater amount.

Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 ) Properties of Solutions Solubility Solubility is the amount of solute that dissolves in a given amount of solvent at a given temperature to produce a saturated solution. Units: grams solute/100 g solvent NaCl: solubility of 36.2 g/ 100 g water at 25 o C Any amount less than that is an unsaturated solution. A solution that contains more solute than it should theoretically is supersaturated.

Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 ) Concentration: Measure of the amount of solute dissolved in a given amount of solvent.  g dm -3 (g/dm 3 ) Is the number of grams of solute dissolved per dm 3 of solution  mol dm -3 (mol/dm 3 ) Molarity (M) is the number of moles of solute dissolved per dm 3 of solution. M = mol dm -3. Use [ ] to signify concentration in molarity.

Solve problems involving concentration, amount of solute and volume of solution. Find the concentration in g dm -3 and mol dm -3 of a solution containing 2.00 g sodium hydroxide in 125 cm 3 of solution. g dm -3  125 cm 3 = dm 3  2.00 g/0.125 dm 3 = 16.0 g dm -3 mol dm -3  2.00 g NaOH = 2.00 g/ g mol -1 =.0500 mol .0500 mol/0.125 dm 3 = mol dm -3

Solve problems involving concentration, amount of solute and volume of solution. Calculate the amount of hydrochloric acid (in mol & g) present in cm 3 of mol dm -3 HCl(aq)?

Solve problems involving concentration, amount of solute and volume of solution. Calculate the amount of hydrochloric acid (in mol & g) present in cm 3 of mol dm -3 HCl(aq)? Molarity = mol dm -3 so Molarity x dm 3 = mol mol = mol dm -3 x dm -3 = mol HCl grams = mol x g mol -1 =.0862 g HCl

Solve problems involving concentration, amount of solute and volume of solution. What volume of a 1.25 mol dm -3 potassium permanganate solution, KMnO 4 (aq), contains 28.6 grams KMnO 4 ?

Solve problems involving concentration, amount of solute and volume of solution. What volume of a 1.25 mol dm -3 potassium permanganate solution, KMnO 4 (aq), contains 28.6 grams KMnO 4 ? Molarity = mol dm -3 so dm -3 = mol/Molarity mol = 28.6 g / g mol -1 =.181 mol dm -3 =.181 mol/1.25 mol dm -3 =.145 dm 3

Solve problems involving concentration, amount of solute and volume of solution. What will be the concentration of the solution formed by mixing 200 cm 3 of 3.00 mol dm -3 HCl(aq) with 300 cm 3 of 1.50 mol dm -3 HCl(aq)?

Solve problems involving concentration, amount of solute and volume of solution. What will be the concentration of the solution formed by mixing 200 cm 3 of 3.00 mol dm -3 HCl(aq) with 300 cm 3 of 1.50 mol dm -3 HCl(aq)? Find total moles.200 dm -3 x 3.00 mol dm -3 =.600 mol HCl.300 dm -3 x 1.50 mol dm -3 =.450 mol HCl Total moles = mol Find total volume:.200 dm dm 3 =.500 dm 3 Concentration: mol/.500 dm 3 = 2.10 mol dm -3

Terms to Know  Element  Compound  Empirical formula  Molecular formula  Structural formula  Percent composition  Percentage yield  Molar mass  Mole  Avogadro’s constant  Chemical equation  Molar ratio  Ionic equation  Solid  Liquid  Gas  Ideal gas equation  Molar volume of a gas  Solute  Solvent  Solution  Concentration