Exam1CKey

Exam1CKey - 1) (18) How many mL of a 1.20 M stock solution...

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(18) How many mL of a 1.20 M stock solution of KCl are required to prepare 800.0 mL of 0.480 M solution? C i V i = C f V f (1.20 M) V i = (0.480 M) (800.0 mL) V i = 320. mL volume = 320. mL 2) (18) How many grams of K 2 SO 4 are required to prepare 500 mL of a solution that is 0.400 M in K 1+ ions? + + 1 4 2 1 K mol 2 SO K mol 1 x L 1 K mol 400 . 0 = 0.200 M K 2 SO 4 (desired conc. of compound) solution L 1 SO K mol 200 . 0 4 2 x 0.5000 L solution = 0.100 mol K 2 SO 4 0.100 mol K 2 SO 4 x 4 2 4 2 SO K mol 1 SO K g 27 . 174 = 17.4 g K 2 SO 4 mass = 17.4 g 3) (18) The mole fraction of ethanol (C 2 H 6 O) in an aqueous solution is 0.20. Its density is 0.896 g/mL. What is the molarity of ethanol in the solution? X = O H mol 0.80 O H C mol 20 . 0 O H C mol 20 . 0 2 6 2 6 2 + b solution L ? O H C mol 20 . 0 6 2 0.20 mol C 2 H 6 O x O H C mol 1 O H C g 46.07 6 2 6 2 = 9.214 g C 2 H 6 O 0.80 mol H 2 O x O H mol 1 O H g 18.02 2 2 = 14.42 g H 2 O 9.214 g C 2 H 6 O + 14.42 g H 2 O = 23.63 g total x solution g 896 . 0 solution
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This note was uploaded on 10/03/2008 for the course CH 201 taught by Professor Warren during the Spring '07 term at N.C. State.

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Exam1CKey - 1) (18) How many mL of a 1.20 M stock solution...

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