CONCENTRATION OF SOLUTIONS

Solute + The amount of solution can be expressed by: - mass m (g, kg) or - volume V (cm 3, mL, dm 3, L, m 3 ) m = V x  - density  (g/cm 3, kg/m 3 ) Solution = Solvent

PERCENT COMPOSITION (by mass) (mass percentage) C  - parts of solute per 100 parts of solution grams of solute per 100 grams of solution Example: 25  aqua NaCl solution. How to prepare solution g? 100 g solution - 25 g NaCl - 75 g H 2 O 400 g solution - x g NaCl - y g H 2 O x = / 100 = 100 g NaCl y = / 100 = 300 g H 2 O V = m /  = 400 / 1.25 = 320 cm 3 (mL)

MOLAR CONCENTRATION (molarity) C M - tells us the number of moles of solute in exactly one liter of a solution (mol/L, mol/1000 mL (cm 3 )) Example: 0.2 M H 2 SO 4 aqua solution. How to prepare 4 Litres? 1 L solution moles 4 L “ - x “ x = / 1 = 0.8 moles H 2 SO 4 1 mol (H 2 SO 4 ) - 98 g/mol 0.8 mol - y g y = 0,8. 98 / 1 = 78.4 g H 2 SO 4

DILUTION OF SOLUTIONS Is used to prepare diluted solutions from concentrated or standard (stock) solutions c 1 V 1 = c 2 V 2 (C M ) c 1 m 1 = c 2 m 2 (C  ) c 1 V 1 d 1 = c 2 V 2 d 2 (C  ) 1 – concentrated (standard) solution, 2 - diluted solution

m t = V. d = = 220 g 1 EXAMPLE.: 200 mL of a solution was prepared by dissolving 17.4 g of K 2 SO 4 in water. What is mass percentage (C  ) and molarity (C M ) of this solution (d=1.1 g/cm 3 ) ? C  - ? 220 g solution g K 2 SO g “ - x g “ x = 7.9 g / 100 g solution = 7.9 

1 EXAMPLE.: 200 mL of a solution was prepared by dissolving 17.4 g of K 2 SO 4 in water. What is mass percentage (C  ) and molarity (C M ) of this solution (d=1.1 g/cm 3 ) ? C M - ? 200 mL solution g K 2 SO mL “ - x g “ x = 87 g/L 1 mol g y mol - 87 g y = 0.5 mol/L

2 EXAMPLE.: If you were going to make 5 L of a 12  (d 2 =1.08 g/cm 3 ) solution of sulfuric acid, how much of concentrated solution (60 , d 1 =1.5 g/cm 3 ) would you need to use? V 1 = c 2 V 2 d 2 / c 1 d 1 = / = = 720 cm 3 60  H 2 SO 4 c 1 m 1 = c 2 m 2 (C  ) c 1 V 1 d 1 = c 2 V 2 d 2 (C  )