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- 1. Concentration of Solute <ul><li>The amount of solute in a solution is given by its concentration . </li></ul>Molarity ( M ) = moles solute liters of solution
- 2. 1.0 L of water was used to make 1.0 L of solution. Notice the water left over.
- 3. PROBLEM: Dissolve 5.00 g of NiCl 2 •6 H 2 O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl 2 •6H 2 O Step 2: Calculate Molarity [ NiCl 2 •6 H 2 O ] = 0.0841 M
- 4. <ul><li>Step 1: Change mL to L. </li></ul><ul><li>250 mL * 1L/1000mL = 0.250 L </li></ul><ul><li>Step 2: Calculate. </li></ul><ul><li>Moles = (0.0500 mol/L) (0.250 L) = 0.0125 moles </li></ul><ul><li>Step 3: Convert moles to grams. </li></ul><ul><li>(0.0125 mol)(90.00 g/mol) = 1.13 g </li></ul>USING MOLARITY moles = M•V What mass of oxalic acid, H 2 C 2 O 4 , is required to make 250. mL of a 0.0500 M solution?
- 5. Learning Check <ul><li>How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? </li></ul><ul><li>1) 12 g </li></ul><ul><li>2) 48 g </li></ul><ul><li>3) 300 g </li></ul>
- 6. The Other Concentration Unit MOLALITY, m m of solution = mol solute kilograms solvent
- 7. Calculating Concentrations <ul><li>Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate molality ethylene glycol. </li></ul>
- 8. Try this molality problem <ul><li>25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution. </li></ul>m = mol solute / kg solvent 25 g NaCl 1 mol NaCl 58.5 g NaCl = 0.427 mol NaCl Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg 0.427 mol NaCl 5 kg water = 0.0854 m salt water
- 9. Preparing Solutions <ul><li>Weigh out a solid solute and dissolve in a given quantity of solvent. </li></ul><ul><li>Dilute a concentrated solution to give one that is less concentrated. </li></ul>
- 10. Dilutions <ul><li>Dilutions are especially important in the lab </li></ul><ul><li>Solutions are shipped in concentrated form, but need to be diluted for use in labs </li></ul><ul><li>Important: </li></ul><ul><ul><li>When you dilute a solution you are NOT changing the number of moles of solute, just volume of solvent </li></ul></ul><ul><ul><li>The concentration of </li></ul></ul><ul><ul><li>the sample is the SAME </li></ul></ul><ul><ul><li>concentration as the </li></ul></ul><ul><ul><li>original solution </li></ul></ul>
- 11. Solving a dilution problem <ul><li>You start with 1 L of a 5 M solution. How would you make this into a 2.5 M solution? </li></ul><ul><li>What if you only wanted 1 L of the 2.5 M solution, and didn’t want to waste any of the 5 M solution? </li></ul><ul><ul><li>We know that any moles brought into the solution stay in the solution. So… </li></ul></ul><ul><ul><ul><li>How many moles do we need in our final solution? </li></ul></ul></ul><ul><ul><ul><ul><li>m = M * V </li></ul></ul></ul></ul><ul><ul><ul><ul><li>m = 2.5 M * 1 L = 2.5 moles </li></ul></ul></ul></ul><ul><ul><ul><li>How what volume of stock solution contains that many moles? </li></ul></ul></ul><ul><ul><ul><ul><li>V = m/M </li></ul></ul></ul></ul><ul><ul><ul><ul><li>V = 2.5 moles /5 M = .5 L </li></ul></ul></ul></ul><ul><ul><ul><ul><li>We need to take out .5 L of stock solution then add water to make 1 L of solution </li></ul></ul></ul></ul>
- 12. M 1 V 1 =M 2 V 2 <ul><li>This is the same process we used on the last slide, just combined into one equation. </li></ul><ul><li>The first side represents the stock solution, the second side represents the diluted solution </li></ul><ul><ul><li>REMEMBER!!!! The volume you solve for is the volume of stock solution you need, this is not the overall answer to the problem!!!! </li></ul></ul>
- 13. <ul><li>Try these! </li></ul><ul><ul><li>If 250 mL of .10 M sodium chloride was diluted to a volume of 750 mL, what will the new concentration be? </li></ul></ul><ul><ul><li>How would you make 50 mL of 3 M solution if you started with 30 M solution? </li></ul></ul>M 1 V 1 =M 2 V 2

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