Molar Concentrations

Molarity is the number of moles of solute that can dissolve in 1 L of solution. Molar concentration (mol/L) = Amount of solute (mol) Volume of solution (L) n C V

Step 1: Find the number of moles of calcium chloride using n=m/M n= 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 28.0 110.98

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 0.252 0.225 L

Solve using 1 step using factor label

Step 1: Find the number of moles using n= cV n= (6.00 mol/L)(0.425L) 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

Solve in 1 step using factor label

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

Factor label method

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

Factor label method

Examples C = ? mNaCl = 14.0 g Vsolution = 250 mL Eg1) 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

Factor label method

Examples Eg 2) 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

Factor label method