Molar Concentrations

Molarity is the number of moles of solute that can dissolve in 1 L of solution. Solute= a solid that will be dissolved Solvent= the liquid doing the dissolving Molar concentration (mol/L) = A 1.00 molar (1.00 M) solution contains 1.00 mol solute in every 1 liter of solution. Units of molarity are: mol/L = M Amount of solute (mol) Volume of solution (L) n C V

Steps involved in preparing solutions from pure solids Calculate the amount of solid required Weigh out the solid Place in an appropriate volumetric flask Fill flask about half full with water and mix/shake. Fill to the mark with water and invert to mix/shake. You should be able to describe this process (including calculating the mass of solid to use) for any solution I specify.

Preparing a 1.0 Molar Solution One liter of a 1.00 M NaCl solution need 1.00 mol of NaCl weigh out 58.5 g NaCl (1.00 mole) and add water to make 1.00 liter (total volume) of solution. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

Steps involved in preparing solutions from pure solids

Step 1: Find the number of moles of calcium chloride using n=m/M n= Ex#1: What is the concentration, in mol/L, of a solution formed by dissolving 28.0g of calcium chloride in enough water to make 225 mL of solution? Step 1: Find the number of moles of calcium chloride using n=m/M n= = = 0.252 mol CaCl2 m M 28.0 g 110.98 g/ mol

Step 2: Use c=n/V to calculate the molar concentration c= = = 1.12 mol/L n V Notice the volume has been converted from mL to L divide by 1000 0.252mol 0.225 L

Step 1: Find the number of moles using n= cV n= (6.00 mol/L)(0.425L) Ex#2: How many grams of sodium nitrate would be needed to make 425 mL of 6.00 mol/L solution? Step 1: Find the number of moles using n= cV n= (6.00 mol/L)(0.425L) = 2.55 mol of sodium nitrate Step 2: Calculate how many grams is needed using m = nM m = (2.55 mol) (85.00 g/mol) = 217 g of NaNO3

nAgNO3 = m/MM C = 0.100 mol/L mAgNO3 = 1.20 g V = ? Ex #3: What final solution volume would be required to prepare A 0.100 mol/L solution of AgNO3(aq) if 1.20 g of the solid salt will be used? MMAgNO3 = 169.9 g/mol C = 0.100 mol/L mAgNO3 = 1.20 g V = ? nAgNO3 = m/MM = 1.20 g 169.9g/mol = 0.00706 mol V = n/C = 0.00706 mol/0.100 mol/L = 0.0706 L x 1000 = 70.6 mL

Ex #4: Conversion of % to mol/L Calculate the concentration, in mol/L, of a 96% solution of sulphuric acid which has a density of 1.84 g/mL Step 1. Calculate the number of moles of sulphuric acid. n= m/M = = 0.98 mol of H2SO4 96g 98.0 g/mol

Step 2: Find the volume of the solution using the mass and the density. Recall: V= m/D = = 54.3 mL = 0.0543 L of solution m 100 g 1.84 g/mL D V

Step 3: Calculate the concentration using c= n/V c = n/V = = 18 mol/L The concentration is 18 mol/L 0.98 mol of H2SO4 0.0543L

Example #5 C = ? mNaCl = 14.0 g Vsolution = 250 mL A Solution of sodium chloride in water contains 14.0 g of NaCl dissolved in 250 mL of solution. What is the molarity of the solution? What do we need to find? Given? MMNaCl = 58.44 g/mol mNaCl = 14.0 g Vsolution = 250 mL C = ? C = n/V n = m/MM = 14.0 g/58.44 g/mol = 0.240 mol NaCl C = n/V = 0.240 mol/0.250 L = 0.960 mol/L

Example#6 What mass of sucrose, C12H22O11, is dissolved in 50.0 mL of a 0.400 mol•L-1 solution of sucrose in water? Given, MMsucrose = 342.3 g/mol C = 0.400 mol/L V = 50.0 mL = 0.0500 L ! nsucrose = C x V = 0.400 mol/L x 0.0500 L = 0.0200 mol sucrose msucrose = n x MM = 0.0200 mol x 342.3 g/mol = 6.85 g of sucrose