II. Solution Concentration (p. 480 – 488)

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

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

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. Mass of Solute x100 Mass of Solution

B. Percent by Mass • How many moles of solute are contained in 343 grams of a 23% aqueous solution of MgCr2O7? 23% 1 mol 343g of solution 100% 240.3 g MgCr2O7 =0.329 mol of MgCr2O7

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. Volume of Solute x100 Volume of Solution

B. Percent by Volume • Determine the percent by volume of toluene (C6H5CH3) in a solution made by mixing 40.0 mL of toluene with 75.0 mL of benzene (C6H6). 40.0 mL toluene + 75.0 mL benzene= 115 mL total solution (40.0 mL toluene / 115 mL solution) 100 = 34.8% toluene

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

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 =.125 mol NaCl 58.44 g NaCl 1 mol NaCl .125 mol NaCl = 7.3 g 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 mol NaF .283 NaF = 0.95 M NaF .25 L sol’n

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

D. Molality • Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 kg water = 3.2m MgCl2

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 water 1.54 mol NaCl 1 kg water 58.44 g NaCl 1 mol NaCl = 45.0 g NaCl

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

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 HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M)V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

500 mL of 1.54M NaCl 500 mLwater 500 mL volumetric flask 500 mL mark 45.0 gNaCl E. Preparing Solutions • 1.54m NaCl in 0.500 kg of water • 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

95 mL of15.8M HNO3 250 mL mark water for safety E. Preparing Solutions • 250 mL of 6.0M HNO3by dilution • measure 95 mL of 15.8M HNO3 • combine with water until total volume is 250 mL • Safety: “Do as you oughtta, add the acid to the watta!” or AA – add acid!

II. Solution Concentration (p. 480 – 488)