# II III I II. Solution Concentration (p. 480 – 488) Ch. 14 – Mixtures & Solutions.

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II III I II. Solution Concentration (p. 480 – 488) Ch. 14 – Mixtures & Solutions

A. Concentration  The amount of solute in a solution  Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

A. Concentration SAWS Water Quality Report - June 2000

B. Percent by Mass  Percent Composition by Mass is the mass of the solute divided by the mass of the solution (mass of the solute plus mass of the solvent), multiplied by 100.solutesolvent Mass of Solute Mass of Solution x100

B. Percent by Mass  How many moles of solute are contained in 343 grams of a 23% aqueous solution of MgCr 2 O 7 ? 343g of solution 23% 100% 1 mol 240.3 g MgCr 2 O 7 =0.329 mol of MgCr 2 O 7

C. Percent by Volume  Percent Composition by Volume is the volume of the solute divided by the volume of the solution (volume of the solute plus volume of the solvent), multiplied by 100.solutesolvent Volume of Solute Volume of Solution x100

B. Percent by Volume  Determine the percent by volume of toluene (C 6 H 5 CH 3 ) in a solution made by mixing 40.0 mL of toluene with 75.0 mL of benzene (C 6 H 6 ). 40.0 mL toluene + 75.0 mL benzene= 115 mL total solution (40.0 mL toluene / 115 mL solution) 100 = 34.8% toluene

C. Molarity  Concentration of a solution total combined volume substance being dissolved

C. Molarity 2M HCl What does this mean?

C. Molarity Calculations  How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L sol’n 0.25 mol NaCl 1 L sol’n = 7.3 g NaCl =.125 mol NaCl 58.44 g NaCl 1 mol NaCl.125 mol NaCl

C. Molarity Calculations  Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g NaF 1 mol NaF 41.99 g NaF.283 NaF.25 L sol’n = 0.95 M NaF =.283 mol NaF

D. Molality mass of solvent only 1 kg water = 1 L water

D. Molality  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 = 3.2 m MgCl 2 0.25 kg water

D. Molality  How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl 58.44 g NaCl 1 mol NaCl

E. Mole Fraction  The number of moles of a component of a solution divided by the total number of moles of all components.molessolution moles Moles A Total Moles = Mole Fraction, Χ

C. Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

C. Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

E. Preparing Solutions  500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL mark 500 mL volumetric flask  1.54m NaCl in 0.500 kg of water

E. Preparing Solutions  250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!” or AA – add acid!

Solution Preparation Mini-Lab  Turn in one paper per team.  Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution.  For each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25m NaCl in 100.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.0M concentrate. (don’t actually prepare this one!)

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