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Solutions - Solutions I Concentration A Definitions-Review...

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1 Solutions I. Concentration. A. Definitions--Review 1. Solution = homogeneous mixture of two or more components. 2. Solvent = component present to largest extent. Phase of solution is same as the phase of the solvent. 3. Solute = minor component of the mixture. a. The solute is dispersed evenly throughout the solvent. b. When the solvent is water the solute is said to be in an aqueous solution. 4. Concentration = number giving the amount of solute dissolved in a given amount solution or dispersed in a given amount of solvent. B. Concentration Units. 1. Review of molarity and mole fraction a. Molarity (M) = moles of solute per liter of solution M = moles of solute liters of solution = mmol solute ml of solution 1) A solution was prepared by dissolving 10.0 g of NaOH (FM = 40.0) in enough water to give 350 mL of solution. Calculate the Molarity of the solution . mol of NaOH = 10.0 g 40 g /mol = 0.250 mol Molarity = 0.250 mol 0.350 L = 0.714 M ( or 0.714 molar) also note mmol of NaOH = 250 mmol, and mL of solution = 350 mL Molarity = 250 mmol / 350 mL = 0.714 M b. Mole fraction (X) 1) The mole fraction of a component , i , in a solution , X i = n i /n total where n i = moles of i n total = total number of moles of all components of the solution. 2) Note that solvent and solute are not used in this expression of concentration. Since the sum of all mole fractions must equal unity, the following holds X i i ! = 1.000 2. Molality (m) = number of moles of solute per kilogram of solvent .
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2 m = moles of solute kg of solvent Suppose that 5.46 g of CH 3 OH (FM = 32) is dissolved in 760 g of water. Calculate the molality of the solution. moles CH 3 OH = 5.46 g 32 g /mol = 0.1706 mol kg of solvent = 0.760 kg molality = 0.1706 mol 0.760 kg = 0.225 molal 3. Interconversion of Concentration Units. a. The calculations of molality and mole fraction require the same type of information, namely the amount of solute (moles) dispersed in a given amount of solvent , expressed in either moles or kilograms. b. The calculation of molarity requires knowledge of the amount (moles) of solute dispersed in a given volume of solution c. One needs to know the density of the solution to interconvert. d. Example: A solution prepared by dissolving 10.0 g of K 2 Cr 2 O 7 (FM = 294) in 90.0 g of water has a density of 1.075 g/mL. Calculate 1) the mole fraction of K 2 Cr 2 O 7 moles of K 2 Cr 2 O 7 = 10.0 g 294 g /mol = 0.0340 mol moles of H 2 O = 90.0 g 18.0 g /mol = 5.00 mol X K 2 Cr 2 O 7 = 0.0340 5.00 + 0.034 = .00675 = 6.75x10 -3 2) the molality of the solution moles K 2 Cr 2 O 7 = 0.0340 mol kg H 2 O = 90.0g 1000 g /kg = 0.0900 kg molality = 0.0340 mol 0.0900 kg = 0.378 molal 3) the Molarity of the solution total mass of solution = 10.0 g + 90.0g = 100.0 g volume of solution = 100.0 g 1.075 g /mL = 93.5 mL = 0.0935 L Molarity = 0.0340 mol 0.935 L = 0.364 molar II. Colligative Properties. Properties of solutions due to their concentrations
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3 A. Vapor Pressure. 1. The vapor pressure of a solution of two volatile liquids will depend on the composition of the mixture and the volatility of the two components.
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