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Explore the concept of solubility and concentration in solutions. Learn about saturated, unsaturated, and supersaturated solutions, as well as the molarity measurement for concentration. Discover how solubility is affected by temperature and pressure, and solve a molarity calculation example.
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DAILY QUESTION November 17, 2008 • What is a saturated solution?
Agenda 11/17/08 • Daily Question • Ch 7 Section 3 Notes • Molarity Worksheet: in-class assignment Assignments: 1. Ch 7 Studyguide due 11/18 Ch 7 Test will be Wednesday, 11/19
Solubility • The maximum amount of a solute that will dissolve in a given quantity of solvent at a given temperature and pressure • 36g of salt can be dissolved in 100ml of water at room temperature • Different substances have different solubilities
Concentration • The quantity of solute dissolved in a given volume of solution • Concentrated: a solution that contains a large amount of solute • Diluted: a solution that contains a small amount of solute
Unsaturated Solutions • Contains less than the maximum amount of solute that will dissolve in the solvent under the same conditions • A solution is unsaturated as long as it is able to dissolve more solute
Saturated Solutions • Can dissolve no more solute • Dissolved solute is in equilibrium with undissolved solute
Supersaturated Solutions • Holds more dissolved solute than is required to reach equilibrium at a given temperature
Solubility of Gases • Depends on pressure • Soda: when opened, the carbon dioxide comes out of solution • Scuba: Nitrogen in blood needs to come out of solution slowly so diver doesn’t get sick
Concentration of Solutions • Molarity is a precise way of measuring concentration Molarity (M) = moles of solute (mol) liters of solution (L) To find moles: mass ÷ molar mass = moles
Finding Molarity What is the molarity of a 0.75 L solution that contains 9.1 g NaCl? NaCl has a molar mass of 58 g/mol? First, Find moles: mass ÷ molar mass = moles Then, Find molarity: moles ÷ volume of solution=molarity 9.1g ÷ 58 g/mol = .1569 mol .1569 mol ÷ 0.75 L = .21 M