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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 soln 0.25 mol NaCl 1 L soln = 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.238 NaF.25 L soln = 0.95 M NaF =.238 mol NaF

Percent Composition by Mass Mass of Solute Mass of Solution x100

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

Percent Composition by Volume 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

Mole Fraction Moles A Total Moles = Mole Fraction, Χ

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 reqd 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

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 Common Terms of Solution Concentration Stock - routinely used solutions prepared in concentrated form. Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl) Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl)

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21 Solutions by Dilution Moles of solute after dilution EQUALS Moles of solute before dilution M 2 x V 2 = M 1 x V 1

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