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Solutions

Solutions. Molarity. Molarity (M). A concentration that expresses the moles of solute in 1 L of solution Molarity (M ) = moles of solute 1 liter solution. Molarity Conversion Factors. A solution is a 3.0 M NaOH. . Write the molarity in the form of conversion factors.

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Solutions

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  1. Solutions Molarity

  2. Molarity (M) A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute 1 liter solution

  3. Molarity Conversion Factors A solution is a 3.0 M NaOH.. Write the molarity in the form of conversion factors. 3.0 moles NaOHand 1 L NaOH soln 1 L NaOH soln 3.0 moles NaOH LecturePLUS Timberlake

  4. Units of Molarity 2.0 M HCl = 2.0 moles HCl 1 L HCl solution 6.0 M HCl = 6.0 moles HCl 1 L HCl solution

  5. Molarity Calculation If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?

  6. Calculating Molarity 1) 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH 40.0 g NaOH 2) 500. mL x 1 L _ = 0.500 L 1000 mL 3. 0.10 mole NaOH= 0.20 mole NaOH 0.500 L 1 L = 0.20 M NaOH

  7. Check for Understanding 1) A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution? 1) 8 M 2) 5 M 3) 2 M Drano

  8. Solution #1 A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution? 2) 5 M M = 2 mole KOH = 5 M 0.4 L Drano

  9. Checking for Understanding 2) A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 0.20 M 2) 5.0 M 3) 36 M

  10. Solution #2 A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 72 g x 1 mole x 1 = 0.20 M 180. g 2.0 L

  11. Making dilutions M1V1 = M2V2 Equation explained…. M1 x V1 = moles before dilution Moles before dilution = moles after dilution M2 x V2 = moles after dilution

  12. Sample problem 4.7 (from textbook) What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution? Known: M1 = 16 M M2 = 0.10 M V2 = 1.5 L Unknown: V1 M1V1 = M2V2 16M x V1= 0.10 M x 1.5 L 0.10 M x 1.5 L = 0.15 mol V1= 0.15 mol 16 M = 0.0094 L or 9.4 mL H2SO4

  13. Practice! What volume of 12.0 M HCl would you need to prepare 200 mL of 3.0 M HCl?

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