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Molarity and Molality

Molarity and Molality. Solutions= Molarity. A Molarity. Molarity = # of moles of a solute per liter of Solution. Solute and Solvent. substance being dissolved. total combined volume. A. Molarity = M. Concentration of a solution. Molarity.

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Molarity and Molality

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  1. Molarity and Molality

  2. Solutions= Molarity A Molarity

  3. Molarity = # of moles of a solute per liter of Solution Solute and Solvent

  4. substance being dissolved total combined volume A. Molarity = M • Concentration of a solution.

  5. Molarity • To make a 0.5-molar (0.5M) solution, first add 0.5 mol ofsolute to a 1-L volumetric flask half filled with distilled water.

  6. Molarity • Swirl the flask carefully to dissolve the solute.

  7. Molarity • Fill the flask with water exactly to the 1-L mark.

  8. A. Molarity (M) 2M HCl What does this mean?

  9. Molarity = M How to REMEMBER hint: Capital M -r…… Liters Mr. Do you want something to drink ?? (liters)

  10. MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molarity Calculations 6.02  1023 (particles/mol) molar mass (g/mol) Molarity (mol/L)

  11. M = Ex: Molarity Calculations • What is the molarity of a solution in which 15.0 g NaCl are dissolved in 400. mL of solution. 15.0 g 1 mol 58.5 g = 0.256 mol NaCl 400 ml = 0.400 L of Solution 0.256 mol .400 L = 0.640 M NaCl

  12. M = Ex. Molarity Calculations • Find the molarity of a 250. mL solution containing 10.0 g of NaF. 10.0 g 1 mol 41.99 g = 0.238 mol NaF 0.238 mol 0.25 L = 0.952M NaF

  13. Ex. Molarity Calculations • How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L 0.25 mol 1 L 58.44 g 1 mol = 7.3 g NaCl

  14. B. Molality Molality

  15. Molality = # of moles of solute dissoved per kg of solvent • Molality = m • m = mole of solute kg of Solvent

  16. Molality = m How to REMEMBER hint: Little l = little m Little Kid (kg)

  17. m = Ex: MolaLity Calculations • How many grams of KI must be dissolved in 500.0 g of H20 to produce a 0.060 molal solution of KI? 1 Kg 1000 g 500 g = 0. 5 kg M = mol kg .5 * .060 = x/.5 Kg Solv. * .5 .03 mol 1 166 g 1mol . KI = 4.98 g KI

  18. moles of solute litersof solution moles of solute m= massof solvent (kg) M = Concentration Units Continued Molarity(M) Molality(m)

  19. Concentrations of Solutions • Water must be tested continually to ensure that the concentrations of contaminants do not exceed established limits. • These contaminants include metals, pesticides, bacteria, and even the by-products of water treatment.

  20. The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. • A concentrated solution contains a large amount of solute. • A dilute solution is one that contains a small amount of solute.

  21. Solution Concentration

  22. Dilution • Dilution:adding solvent to decrease the concentration of a solution • The amount of solute stays the same, but the concentration decreases.

  23. M1 x V1 = M2 x V2 Dilution Formula M1 = concentrated solution (Molarity) V1 = volume of concentrated solution M2 = dilute solution (Molarity) V2 = volume of dilute solution • In dilution we take a certain number of moles of solute and dilute to a bigger volume.

  24. Making Dilutions • The total number of moles of solute remains unchanged upon dilution, so you can write this equation. • M1 and V1 = molarity and volume of the concentrated solution, and • M2 and V2 = molarity and volume of the diluted solution.

  25. Making a Dilute Solution

  26. Make 800.0 mL of a 0.25M solution of HCl. The only availabe HCl is concentrated (12M) Step l: M1 M1 x V1 = M2 x V2 M2 = 0.25M HCl = 12.0 M HCl V1 = ? V2= 800.0mL Step 2: (12.0 M) (V1) = (0.25M) (800.0 mL) V1 = (0.25M HCl) (800.0 mL )(12.0 M HCl) V1 = 16.67 mL

  27. V1 = 16.67 mL • This 16.67 mL is the amount of concentrated acid we will take out. • We still have to make 800 mL of 0.25M • 800mL-16.67mL =783.3 mLof solution must bewater • So we would place16.67 mL of HCl into 783.3 mL of water

  28. To prepare 100 ml of 0.40M MgSO4 from a stock solution of 2.0M MgSO4, a student first measures 20 mL of the stock solution with a 20-mL pipet. • She then transfers the 20 mL to a 100-mL volumetric flask. • c) Finally she carefully adds water to the mark to make 100 mL of solution.

  29. Mass Percent • The concentration of a solution in percent can be expressed : • as the ratio of the mass of the solute or solvent to the mass of the solution.

  30. x 100% mass of solute x 100% = mass of solution mass of solute mass of solute + mass of solvent Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass % by mass =

  31. What is the percent by mass of a solution containing 2.3 g of ethanol (C2H5OH) dissolved in 10.0 g of water? % Step 1: Find the mass of the solution 10.0 g H20 + 2.3 g (C2H5OH) = 12.3 g Step 2: Divide mass of solute by mass of solution and multiply by 100% %mass = 2.3 g (C2H5OH) X 100% = 19% 12.3 g solution

  32. Concentration in Percent • (Volume/Volume) • Isopropyl alcohol (2-propanol) is sold as a 91% solution. • This solution consist of 91 mL of isopropyl alcohol mixed with enough water to make 100 mL of solution.

  33. Saline Solution Concentration Percentage • Mass percent = grams of solute per 100 g of solution • 0.9% NaCl has 0.9 g of NaCl in every 100 g of solution • Mass of solution = mass of NaCl + mass of water • Divide the mass of solute by the mass of solution and multiply by 100% to get mass percent.

  34. Key Concepts: Solutions • Solutions are homogeneous mixtures. • Many reactions occur in aqueous solutions because ions are available to react. • Solvent: the substance present in the highest percentage • Solute: the dissolved substance, which is present in lesser amount • Molality is expressed as Moles/kg of solution • Molarity is expressed as Moles/Liter of solution.

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