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1 Concentration Measurement: Molarity Mr. ShieldsRegents Chemistry U12 L02

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2 Molarity Let’s start our discussion of solution concentration Measurement with Molarity: the most common solution concentration measurement is defined as follows: M = moles of solute LITERS of solution LITERS of solution Which means Molarity = # of moles per liter of solution MOLARITY Capital M

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3 Molarity Let’s Consider the following Problem: What is the molarity of a solution in which 50g of CuSO 4 is dissolved in water to make a 2 liter solution. Step 1. What’s the formula for MOLARITY? M = moles of solute liters of solution

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4 Molarity Step 2. We need to determine how many moles we have in 50g of CuSO 4. How do we do that? Remember the mole Hole? - 1. Determine the molar mass of CuSO 4 (159.5g/mol) - 2. Calculate the # of moles of CuSO 4 Since we’re going into the mole hole we need Since we’re going into the mole hole we need to DIVIDE # of grams by molar mass to DIVIDE # of grams by molar mass (50/159.5) = 0.31 moles

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5 Molarity Step 3. Calculate the Molarity of the solution using the formula for MOLARITY M = moles of Solute liters of solution How many moles do we have? How many liters do we have? So M = 0.31 moles 2 liters M = 0.155 mol/liter 0.31 mol 2 L

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6 Making a 1M NaCl Solution 1. Weigh out and add 1 mole of NaCl to a volumetric Flask 2. Add a small amount of solvent to dissolve the salt 3. Once the solute is dissolved add additional solvent up to the 1 liter mark

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7 Problem What is the molarity of a solution made by dissolving 2.1 liters of NH 3 (g) in sufficient water to make 2 liters of solution? 1.Calculate how many moles of ammonia there are # of moles = volume = 2.1 liters = 0.094 mol molar vol 22.4 l/mol What do we need to begin calculating the Molarity of the solution?

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8 Problem Step 2. Calculate the molarity of the solution using the molarity equation M = # moles = 0.094 mol = 0.047M # liters of sol’n 2 liters We now know the number of moles of NH 3 we have (0.094) so …

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9 conc dil Molarity by dilution Now what if we had a solution already prepared but Wanted to decrease the molarity to some lower precise Value. How could we use an already prepared higher molarity solution to accomplish this? Note that it is only possible to go from a more conc. (higher molarity) solution to a less conc. (lower molarity) solution.

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10 Molarity by dilution Recall that Molarity (M) = # moles of solute # liters of solution So we need to dilute the solution by adding solvent But how much more solvent do we need to add? To make something more dilute we’re going to need to add more solvent to the solution.

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11 Molarity by dilution When we add more solvent to a solution do we change the number of moles of solute in the solution? NO! But what does change? We change the # of moles per unit volume of Solvent because the same # of moles are present but now there’s more liquid

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12 Molarity by Dilution Let’s call the Molarity of the original solution M1 and the New Molarity M2 The number of liters of the original solution and the new Solution will then be designated by V1 and V2 So the molarity of the old solution is… M1 = #moles of solute= #moles #liters of solution V1 So… #moles = M1 x V1 Let’s see how this works …

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13 Molarity by Dilution The Molarity of the new solution will then be … M2 = #moles of solute #liters of solution # moles = M2 x V2 Remember that the # of moles does not change when We dilute solutions.

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14 Molarity by dilution So… M1 x V1 = M2 x V2 If we know three values we can calculate the 4 th. Let’s see how we use this relationship

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15 Dilution Problem How much of a 3M solution of NiCL 2 is needed to Prepare 100 ml of a 1.75M solution of NiCl 2 ? OK how do we solve this? What’s the equation we’re going to use? Right!M1V1 = M2V2 M1 = 3M;M2 = 1.75M;V2 = 0.1L (100ml) So … 3 x V1 = 1.75 x 0.1 V1 = 0.058Lor 58ml

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16 Molarity by dilution How do we make 5L of A 1.5M sol’n of KCL From 12.0M stock sol’n? M1V1 = M2V2 12 x V1 = 1.5 x 5 12V1= 7.5 V1 = 7.5/12 = 0.625L Add 4.375L H 2 0 to a 0.625L of a 12M sol’n

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