Other units of concentration Dilution Equation MV MV

Other units of concentration

Dilution Equation MV = MV n Molarity (volume) before dilution = molarity (volume) after dilution n How many liters of 12 M H 2 SO 4 do you need to make 1. 2 L of. 75 M? n 12 M (V) =. 75 M (1. 2 L) n V =. 075 L (75 m. L) n

Mass Percent Mass percent = mass of solute x 100 n mass of solution n Or = grams of solute x 100 n grams of solute + grams of solvent n

Problem 35 g of Na. Cl is dissolved in 115 g of water, what is the mass percent? What is the molarity if the final solution has a density of 1. 1 g/m. L? n 35 g / (35 g + 115 g) x 100 n 23 % n 35 g x 1 mol/ 58. 44 g=. 5989 mol Na. Cl n 150 g x 1 m. L / 1. 1 g = 136. 36 m. L =. 13636 L n M =. 5989 mol /. 13636 L n M = 4. 4 M n

Convert 1. 2 M Cu. SO 4 solution to mass percent, if the solution has a density 1. 1 g/m. L. n 1. 2 M= 1. 2 mol Cu. SO 4 / 1 L solution n 1. 2 mol x 159. 62 g / 1 mol =191. 544 g n 1 L = 1000 m. L x 1. 1 g/1 m. L = 1100 g of solution n n Mass percent = 191. 544 g / 1100 g x 100 = 17 %

More practice n 55 g of Ca. Cl 2 is dissolved in 115 g of water, what is the mass percent? What is the molarity if the final solution has a density of 1. 1 g/m. L?

More n Convert 1. 9 M Ca(NO 3)2 solution to mass percent, if the solution has a density 1. 3 g/m. L.

Parts per thousand/million etc. Mass percent can also be called parts per hundred (although it never is) n Parts per thousand is the same as mass percent except instead of multiplying by a 100 you multiply by 1000. n Parts per million is multiplied by 1, 000 n Pollen counts are normally reported in this n

Mole Fraction

Mole Fraction or molar fraction is represented by the Greek letter chi, n = mole solute/mole solution n

Convert a 2. 3 M of water solution of aluminum nitrate Al(NO 3)3 to mole fraction of aluminum nitrate if the density of the solution is 1. 14 g/m. L. n 2. 3 M = 2. 3 mol Al(NO 3)3 /1 L solution n 2. 3 mol x 213. 01 g/ 1 mol = 489. 923 g Al(NO 3)3 n 1 L = 1000 m. L x 1. 14 g/1 m. L = 1140 g solution n 1140 g - 489. 923 g = 650. 077 g water n 650. 077 g x 1 mol /18. 016 g = 36. 0833 mol n = 2. 3 mol/ (2. 3+36. 0833) mol =. 060 n

Another Convert a 1. 4 M solution of strontium sulfate, Sr. SO 4, in water to mole fraction if the density of the solution is 1. 24 g/m. L. n 1. 4 M = 1. 4 mol Sr. SO 4 /1 L solution n 1. 4 mol x 183. 68 g/ 1 mol = 257. 152 g Sr. SO 4 n 1 L = 1000 m. L x 1. 24 g/1 m. L = 1240 g solution n 1240 g – 257. 152 g = 982. 848 g water n 982. 848 g x 1 mol /18. 016 g = 54. 554 mol n = 1. 4 mol/ (1. 4+54. 554) mol =. 025 n

Other way Convert a mole fraction. 032 of a K 2 CO 3 solution in water to molarity if the density is 1. 08 g/m. L. n 0. 032 K 2 CO 3 =. 032 mol /1 mol solution n Mol H 2 O = 1. 00 -. 032 =. 968 mol n. 032 mol K 2 CO 3 x 138. 21 f g/mol = 4. 422 g n. 968 mol H 2 O x 18. 016 g/mol = 17. 439 g n Total solution = 17. 439 + 4. 422 = 21. 86 g n 1. 08 g/m. L = 21. 86 g/ V, V = 20. 24 m. L n M =. 032/. 02024 = 1. 6 M K 2 CO 3 n