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Understanding Solutions: Solvent, Solute, and Concentration

Explore the concept of solutions, including the roles of solute and solvent, solution concentration, and dilution equations. Practice molarity calculations and solve dilution problems. Get ready to ace your chemistry homework!

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Understanding Solutions: Solvent, Solute, and Concentration

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  1. Solutions Chapter 14

  2. What are Solutions? • Solution : a uniform (homogeneous) mixture that may contain solids, liquids or gases. • Parts of a solution: • Solute: What gets dissolved in a solution – usually present in smaller amounts • Solvent: What does the dissolving in a solution – usually present in larger amounts – WATER is the “UNIVERSAL SOLVENT”

  3. 14.2 - Solution Concentration • Concentration: a measure of how much solute is dissolved in a specific amount of solvent or solution. • We will express concentration in units of Molarity (which will be explored at a later time) • Example: a solution with a concentration of 1.5 M is said to be a 1.5 “Molar” solution

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

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

  6. Molarity 2M HCl What does this mean?

  7. M = 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 = 0.95M NaF

  8. Solution Concentration (cont’d) • A “Concentrated” solution contains a large amount of solute relative to its solubility • A “dilute” solution contains a small amount of solute relative to its solubility

  9. The Dilution Equation M1V1 = M2V2 M1 = initial molarity (“stock solution”) V1 = initial volume (L) M2 = final (desired) molarity V2 = final volume (L) This equation is used when you have a “stock solution” of higher molarity than you need and you need to dilute it to a lower molarity by adding additional solvent.

  10. Dilution Equation (cont’d) What volume of 3.00M KI stock solution would you use to make 0.300 L of a 1.25M KI solution?

  11. DILUTION PROBLEMS How much concentrated 18 M sulfuric acid is needed to prepare 250 mL of a 6.0 M solution? How much concentrated 12 M hydrochloric acid is needed to prepare 100 mL of a 2.0 M solution? What volume of 3.0 M nitric acid could be made from 25 mL of a 15 M stock solution? If I add 50 mL of water to 130 mL of 12 M hydrochloric acid, what is the new, diluted concentration? If I add 100 mL of water to 100 mL of 6 M nitrous acid, what is the new, diluted concentration?

  12. HOMEWORK • COMPLETE THE LAST 5 DILUTION PROBLEMS AS HOMEWORK • FOUND UNDER THE HOMEWORK TAB • DUE TOMORROW AT THE START OF CLASS: • THE 5 QUESTIONS DONE IN CLASS • THE 5 HOMEWORK QUESTIONS • FOR A TOTAL OF 10 DILUTION PROBLEMS

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