# Concentration of Solution Solvent Solute Concentration of Solution Concentration of Solution Solvent Solute Concentration of Solution Moles of solute Mol (M) = = Liter of solution Molarity L Parts amount of solute (g or ml) 2 6 9 (10 ) or (10 ) or (10 ) = ratio amount of solution (g or ml) Mole Fraction()= Total moles of solution Moles of solute

Molality (m) = Moles of solute Kilograms of solvent Molarity NaCl Molarity Example Problem 1 12.6 g of NaCl are dissolved in water making 344mL of solution. Calculate the molar concentration. moles solute M= L solution 1molNaCl 12.6 g NaCl 58.44 gNaCl = 1L 344 mL solution 1000mL

= 0.627 M NaCl Molarity NaCl Molarity Example Problem 2 How many moles of NaCl are contained in 250.mL of solution with a concentration of 1.25 M? moles solute M= L solution 1L 250. mL = 0.250 L solution 1000mL Volume x concentration therefore the solution contains 1.25 mol NaCl 1 L solution = moles solute 1.25 mol NaCl 0.250 L solution = 0.313

1 L solution mol NaCl Molarity NaCl Molarity Example Problem 3 What volume of solution will contain 15 g of NaCl if the solution concentration is 0.75 M? moles solute M= L solution therefore the solution contains 1 mol NaCl 15 g NaCl = 0.257 mol 58.44 g NaCl 0.75 mol NaCl 1 L solution moles solute concentration = volume solution 1 L solution 0.257 mol NaCl =

0.75 mol NaCl 0.34 L solution % Concentration mass solute % (w/w) = mass solution % (w/v) = x 100 mass solute volume solution x 100 volume solute volume solution x 100 % Mass (v/v) =and volume units must match. (g & mL) or (Kg & L) % Concentration Example Problem 1 What is the concentration in %w/v of a solution containing 39.2 g of potassium nitrate in 177 mL of solution?

39.2 g mass solute 100 = 22.1 % w/v 100 % (w/v) = volume solution 177 mL Example Problem 2 What is the concentration in %v/v of a solution containing 3.2 L of ethanol in 6.5 L of solution? volume solute 100 % (v/v) = volume solution 3.2 L 100 6.5 L = 49 % v/v % Concentration Example Problem 3 What volume of 1.85 %w/v solution is needed to provide 5.7 g of solute? 1.85 g solute % (w/v) = 100 mL solution We know: g solute and

We want to get: g solute mL solution 100 mL solution 5.7 g solute 1.85 g solute g solute concentration mL solution = 310 mL Solution = volume solution Parts per million/billion (ppm & ppb) mass solute 106 volume solution or mg = ppm L mass solute 109

ppb = volume solution or g L ppm = Mass and volume units must match. (g & mL) or (Kg & L) = ppb AND For very low concentrations: ng parts per trillion L = ppt ppm & ppb Example Problem 1 An Olympic sized swimming pool contains 2,500,000 L of water. If 1 tsp of salt (NaCl) is dissolved in the pool, what is the concentration in ppm? 1 teaspoon = 6.75 g NaCl

g solute ppm = 106 mL solution 6.75 g 6 ppm = 10 mL 2.5106 L 1000 1 L ppm = 0.0027 or mg solute ppm = L solution ppm = mg 6.75 g 1000 1 g 2.5106 L ppm = 0.0027 ppm & ppb Example Problem 2 An Olympic sized swimming pool contains 2,500,000 L of water. If 1 tsp of salt (NaCl) is dissolved in the pool, what

is the concentration in ppb? 1 teaspoon = 6.75 g NaCl g solute ppb = 109 mL solution 6.75 g 9 ppb = 10 mL 2.5106 L 1000 1 L ppb = 2.7 or g solute ppb = L solution ppb = 6.75 g 106 mg 1 g 2.5106 L ppb = 2.7

Mole Fraction A A B A A B A A B A B B A A Mole Fraction () moles of A A A = sum of moles of all components moles of B B B = sum of moles of all components Since A + B make up the entire mixture, their mole

fractions will add up to one. A A + + B B A B 1.00 Mole Fraction Example Problem 1 In our glass of iced tea, we have added 3 tbsp of sugar (C12H22O11). The volume of the tea (water) is 325 mL. What is the mole fraction of the sugar in the tea solution? (1 tbsp sugar 25 g) First, we find the moles of both the solute and the solvent. 1 mol C12 H 22 O11 75.g C12 H 22O11 = 0.219 mol 342 g C12 H 22 O11 1 mol H 2 O 325mL H 2 O = 18.1 mol

18.0 g H O 2 Next, we substitute the moles of both into the mole fraction equation. 0.219 mol sugar moles solute = sugar = total moles solution (0.219 mol + 18.1 mol) 0.012 Mole Fraction Example Problem 2 Air is about 78% N2, 21% O2, and 0.90% Ar. What is the mole fraction of each gas? First, we find the moles of each gas. We assume 100. grams total and change each % into grams. 1 mol N 2 78g N 2 = 2.79 mol 28 g N 2 1 mol O 2 21g O 2

= 0.656 mol 32 g O 2 Next, we substitute the moles of each into the mole fraction equation. 1 mol Ar 0.90g Ar = 0.0225 mol 40. g A r = moles N 2 = N2 total moles 2.79 mol N 2 (2.79 + 0.656 + 0.0225) 0.804 =

moles O 2 = O2 total moles 0.656 mol O 2 (2.79 + 0.656 + 0.0225) 0.189 = Ar = moles Ar total moles 0.0225 mol Ar (2.79 + 0.656 + 0.0225) 0.00649 Molal (m) Example Problem 1 If the cooling system in your car has a capacity of 14 qts, and you want the coolant to be protected from freezing down to -25F, the label says to combine 6 quarts of antifreeze with 8 quarts of water. What is the molal concentration of the antifreeze in the mixture? mol solute m= Kg solvent m=

antifreeze is ethylene glycol C2H6O2 1 qt antifreeze = 1053 grams 1 qt water = 946 grams 1053 g C 2 H 6O 2 1mol C2 H 6O 2 6 Qts 62.1 g C H O 1 Qt C H O 2 6 2 2 6 2 946 g H 2O 8 Qts 1 Qt H O 2 1 Kg 1000 g = 13 m