# II III I II. Concentration Ch. 13 & 14 - Solutions.

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II III I II. Concentration Ch. 13 & 14 - Solutions

What is different between the glasses of Kool-aid?

Solution concentration can be described generally  Dilute - reduced in strength, weak, watered down.  Concentrated – stronger, pure. Has less water.

What is the problem with just using dilute and concentrated as descriptions of the solution concentration?

Is solution B dilute or concentrated?  The terms dilute and concentrated are relative.  Scientists need a more precise way of referring to the concentration of a solution. ConcentratedDilute Solution A Solution B Solution C

Solution concentration can be described specifically  Do you remember the “mole” from Stoichiometry?  What is a mole?  How might you use it to describe the concentration of a solution?

A. Concentration  The amount of solute in a solution.  Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

Molarity The ratio of the moles of solute to the volume of solution in liters. Molarity (M) = Moles of solute Volume in Liters of Solution

How to read Molarity  6.0 M NaCl  Read: “6 molar solution of NaCl”  Can be abbreviated 6M solution  You must be careful to label the molarity with a capital M so that it is not confused with m for molality.

How to make a 6M NaCl solution using molarity (a) Add 6 moles NaCl to the volumetric flask. How would you measure that? 6 molesNaCl 58.443 g NaCl 1 mole NaCl = 351 g NaCl

How to make a 6M NaCl solution using molarity (b) Add distilled or deionized H 2 O to dissolve and mix the NaCl (c) Fill the flask with dH 2 O until you reach the 1000mL line.

Types of Calculations with Molarity: 1. Finding concentration of a solution. 2. Finding the mass of solute needed. 3. Finding the volume of solution made.

Finding Concentration  Antifreeze is a solution of ethylene glycol, C 2 H 6 O 2 in water. If 4.50 L of antifreeze contains 27.5 g of ethylene glycol, what is the concentration of the solution? 27.5 g C 2 H 6 O 2 62.08 g C 2 H 6 O 2 1 mol C 2 H 6 O 2 4.5 L = 0.0984 mol/L or 0.0984 M C 2 H 6 O 2

Finding Mass  What mass of sodium carbonate, Na 2 CO 3, is present in 50 ml of a 0.750M solution? 50 ml 1 L 1000 mL 0.750 mol 1 L 1 mol Na 2 CO 3 105.99 g Na 2 CO 3 = 3.97 g Na 2 CO 3 Conversion Factor

Finding Volume  What volume of 1.50 mol/L HCl solution contains 10.0 g of hydrogen chloride? 10.0 g HCl 1 mol HCl 36.46 g HCl 1.50 mol 1 L = 0.183 L or 183 mL Conversion Factor

Practice Problems

1. A 0.750 L aqueous solution contains 90.0 g of ethanol, C 2 H 5 OH. Calculate the molar concentration of the solution in mol/L.

Practice Problem 2. What mass of NaCl are dissolved in 152 mL of a solution if the concentration of the solution is 0.364 M?

Practice Problem 3. What mass of dextrose, C 6 H 12 O 6 is dissolved in 325 mL of 0.258 M solution?

Practice Problem 4. A mass of 98 g of sulfuric acid, H 2 SO 4, is dissolved in water to prepare a 0.500 M solution. What is the volume of the solution?

Practice Problem 5. A solution of sodium carbonate, Na 2 CO 3, contains 53.0 g of solute in 215 mL of solution. What is its molarity?

Practice Problem 6. What is the molarity of a solution of HNO3 that contains 12.6 g of solute in 5.00 L of solution?

B. Molality mass of solvent only 1 kg water = 1 L water

B. Molality  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 = 3.2 m MgCl 2 0.25 kg water

B. Molality  How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl 58.44 g NaCl 1 mol NaCl

Practice Problem What is the molality of a solution made from 2.4 moles of NaCl and 0.80 kg of water?

Practice Problem What is the molality of a solution made from 63 g of HNO 3 in 0.50 kg of water?

Practice Problem How much water is needed to make a 0.50 m solution from 3.2 g of NaCl?

Practice Problems What mass of CH 3 OH is needed to add to 1.20 kg of water to make a 3.00 m solution?

C. Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

C. Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

Dilution Practice Water is added to 200. mL of a 2.0 M solution of CaCl 2 to increase the volume of the solution to 400. mL. What is the new concentration?

Dilution Practice To what volume must 1.0 L of a 6.0 M solution of HCl be diluted in order to prepare a 0.2 M solution?

D. Preparing Solutions  500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL mark 500 mL volumetric flask  1.54m NaCl in 0.500 kg of water

D. Preparing Solutions  250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!”

Solution Preparation Lab  Turn in one paper per team.  Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution.  For each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25m NaCl in 100.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.1M concentrate.

500.O ml of 0.50 M NaCl  Remember: 0.50 M = 0.50 mole/1 Liter  Convert 0.50 moles to grams of NaCl  0.50 mole NaCl = 29.23 grams but this is in 1 liter  Figure out how much you need for 0.1 L  Set up a proportion. 29.23g = x 1 L 0.5 L  X = 14.62 g, measure this amount on a balance and add to a volumetric flask.  Then fill with distilled water to the 500 ml mark.

0.25 m NaCl in 100 ml of water  Remember: 0.25 m = 0.25 moles/1kg of water  Convert 0.25 moles to grams.  0.25 moles = 14.61 grams – this is in 1 kg of water.  Figure out how much you need for 0.100 kg of water.  1kg = 1L  Set up a proportion. 14.61 g = x 1 kg 0.100 kg  X = 1.461 g, measure this amount on a balance and add to a flask.  Then measure out 100 ml of water in a graduated cylinder. Add this volume of water to the flask containing the NaCl.

How do you make a 100.0 mL solution with a concentration of 3.0 M HCl from 12.1M concentrate  This is a dilution problem because you have 2 different concentrations.  Solve for the first volume: (12.1)(V 1 ) = (3.0)(0.100)  V 1 = 0.025 L or 25 mL  Measure out 25 mL of HCl in a graduated cylinder  In a 100 mL volumetric flask, add about 50 mL of water then slowly pour the 25 mL of HCl into the flask. Then add the remaining volume of water to the 100 ml mark.

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