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1 Chapter 7 Solutions 7.5 Molarity and Dilution

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2 Molarity (M) Molarity (M) is a concentration term for solutions. gives the moles of solute in 1 L solution. moles of solute liter of solution

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3 Preparing a 1.0 Molar Solution A 1.00 M NaCl solution is prepared by weighing out 58.5 g NaCl (1.00 mole) and adding water to make 1.00 liter of solution. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

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4 What is the molarity of 0.500 L NaOH solution if it contains 6.00 g NaOH? STEP 1 Given 6.00 g NaOH in 0.500 L solution Need molarity (mole/L) STEP 2 Plan g NaOH mole NaOH molarity Calculation of Molarity

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5 Calculation of Molarity (cont.) STEP 3 Conversion factors 1 mole NaOH = 40.0 g 1 mole NaOH and 40.0 g NaOH 40.0 g NaOH 1 mole NaOH STEP 4 Calculate molarity. 6.00 g NaOH x 1 mole NaOH = 0.150 mole 40.0 g NaOH 0.150 mole = 0.300 mole = 0.300 M NaOH 0.500 L 1 L

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6 What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO 3 ? 1) 0.557 M 2) 1.44 M 3) 1.71 M Learning Check

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7 3) 1.71 M 46.8 g NaHCO 3 x 1 mole NaHCO 3 = 0.557 mole NaHCO 3 84.0 g NaHCO 3 0.557 mole NaHCO 3 = 1.71 M NaHCO 3 0.325 L Solution

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8 What is the molarity of 225 mL of a KNO 3 solution containing 34.8 g KNO 3 ? 1)0.344 M 2)1.53 M 3)15.5 M Learning Check

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9 2)1.53 M 34.8 g KNO 3 x 1 mole KNO 3 = 0.344 mole KNO 3 101.1 g KNO 3 M = mole = 0.344 mole KNO 3 = 1.53 M L 0.225 L In one setup: 34.8 g KNO 3 x 1 mole KNO 3 x 1 = 1.53 M 101.1 g KNO 3 0.225 L Solution

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10 Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions. TABLE 7.8

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11 Calculations Using Molarity How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution? STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl Need Grams of KCl STEP 2 Plan L KCl moles KCl g KCl

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12 Calculations Using Molarity STEP 3 Conversion factors 1 mole KCl = 74.6 g 1 mole KCl and 74.6 g KCl 74.6 g KCl 1 mole KCl 1 L KCl = 0.720 mole KCl 1 L and 0.720 mole KCl 0.720 mole KCl 1 L STEP 4 Calculate grams. 0.125 L x 0.720 mole KCl x 74.6 g KCl = 6.71 g KCl 1 L 1 mole KCl

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13 How many grams of AlCl 3 are needed to prepare 125 mL of a 0.150 M solution? 1) 20.0 g AlCl 3 2) 16.7g AlCl 3 3) 2.50 g AlCl 3 Learning Check

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14 Solution 3) 2.50 g AlCl 3 0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl 3 1 L 1 mole

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15 How many milliliters of 2.00 M HNO 3 contain 24.0 g HNO 3 ? 1) 12.0 mL 2) 83.3 mL 3) 190. mL Learning Check

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16 24.0 g HNO 3 x 1 mole HNO 3 x 1000 mL = 63.0 g HNO 3 2.00 mole HNO 3 Molarity factor inverted = 190. mL HNO 3 Solution

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17 D ilution In a dilution water is added. volume increases. concentration decreases.

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Mix The amount of solute in the concentrated solution = amount of solute in the diluted solution 3 containers of H 2 O 1 container orange juice concentrate +

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19 Comparing Initial and Diluted Solutions In the initial and diluted solution, the moles of solute are the same. the concentrations and volumes are related by the following equations: For percent concentration: C 1 V 1 = C 2 V 2 initial diluted For molarity: M 1 V 1 = M 2 V 2 initial diluted

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20 Dilution Calculations with Percent What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution? Prepare a table: C 1 = 14.0% (m/v)V 1 = 25.0 mL C 2 = 2.00% (m/v)V 2 = ? Solve dilution equation for unknown and enter values: C 1 V 1 = C 2 V 2 V 2 = V 1 C 1 = (25.0 mL)(14.0%) = 175 mL C 2 2.00%

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21 Learning Check What is the percent (% m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

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22 Solution What is the percent (%m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL? Prepare a table: C 1 = 9.00 %(m/v)V 1 = 10.0 mL C 2 = ?V 2 = 60.0 mL Solve dilution equation for unknown and enter values: C 1 V 1 = C 2 V 2 C 2 = C 1 V 1 = (10.0 mL)(9.00%) = 1.50% (m/v) V 2 60.0 mL

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23 Dilution Calculations with Molarity What is the molarity (M) of a solution prepared by diluting 0.180L of 0.600 M HNO 3 to 0.540 L? Prepare a table: M 1 = 0.600 MV 1 = 0.180 L M 2 = ?V 2 = 0.540 L Solve dilution equation for unknown and enter values: M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 = (0.600 M)(0.180 L) = 0.200 M V 2 0.540 L

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24 Learning Check What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? 1) 27.0 mL 2) 60.0 mL 3) 90.0 mL

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25 Solution What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? Prepare a table: M 1 = 1.80 MV 1 = 15.0 mL M 2 = 0.300MV 2 = ? Solve dilution equation for V 2 and enter values: M 1 V 1 = M 2 V 2 V 2 = M 1 V 1 = (1.80 M)(15.0 mL) = 90.0 mL M 2 0.300 M

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