Assignment Solutions CHEM 110- Fall 2011 1. Express the following ...

3. For a solution of acetic acid (CH 3 COOH) to be called “vinegar”, it must contain
5.00% acetic acid by mass (mass percent %w/w). Vinegar is a solution of acetic
acid in water. What is the molarity of acetic acid in the vinegar? The density of
vinegar is 1.006 g/ml.
CH 3 COOH mass = 5.00 grams in 100 grams of solution for 5.00% (w/w).
Mole CH 3 COOH = 5.00 g x
1mole
60 grams
= 0.0833 mole CH 3 COOH
Volume of solution = 100 grams of solution x
1ml
1.006 g
= 99.4 mL
Molarity =
mole of solute
volume of solution (L)
=
0.0833 mole Acetic acid
0.0994 L
= 0.838 M
4. How would you prepare 465 mL of 0.3550 M potassium dichromate (K 2 Cr 2 O 7 )
solution starting with
a) solid potassium dichromate
Determine mole of K 2 Cr 2 O 7 required = 0.465 L x 0.3550
mole = 0.165 mole
L
Determine mass of K 2 Cr 2 O 7 = 0.165 mole x
gram
2 .94 = 48.5 grams
mole
Measure and dissolve 48.5 gram in a total solution volume of 465 ml.
b) 0.750 M potassium dichromate solution.
M 1 V 1 = M 2 V 2
M 1 = 0.750 M : V 1 = ? M 2 = 0.3550 M; V 2 = 465 mL
(0.750 M)(V 1 ) = (0.3550 M) (465 mL)
V 1 =
(0.3550 M)(465 mL)
0.750 M
= 220.1 mL
Measure 220.1 mL of 0.750 M K 2 Cr 2 O 7 into an appropriate size flask then dilute
with water until the total solution volume is 465 mL.