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

Assignment Solutions CHEM 110- Fall 2011
1. Express the following in %w/v and molarity.
a) 5.62 g of NaHCO 3 dissolved in water to make a 250 mL solution
w
% =
v
mass of solute (g)
5.62 g
x 100 %w/v NaHCO 3 = x 100 = 2.25 %
volume of solution (ml)
250 ml
Molarity =
mole of solute (g)
volume of solution (L)
1mole
5.62 g x
84 g
Molarity NaHCO 3 =
0.250 L
=0.268 M
b) A 500.0 mL solution containing 184.6 g K 2 Cr 2 O 7
w
% =
v
mass of solute (g)
184.6 g
x 100 %w/v K 2 Cr 2 O 7 = x 100 = 36.9 %
volume of solution (ml)
500 ml
Molarity =
mole of solute (g)
volume of solution (L)
1mole
184.6 g x
294 g
Molarity K 2 Cr 2 O 7 =
0.500 L
=1.26 M
c) A 250 mL solution containing 17.6 grams of ascorbic acid
w
% =
v
mass of solute (g)
17.6 g
x 100 %w/v ascorbic acid = x 100 = 7.04 %
volume of solution (ml)
250 ml
Molarity =
mole of solute (g)
volume of solution (L)
Molarity ascorbic acid =
1mole
17.6 g x
176 g
0.250 L
=0.400 M
d) 4.6 gram of ethanol diluted to 250 mL.
w
% =
v
mass of solute (g)
4.6 g
x 100 %w/v ethanol = x 100 = 1.84 %
volume of solution (ml)
250 ml
Molarity =
mole of solute (g)
volume of solution (L)
Molarity ethanol =
1mole
4.6 g x
46 g
0.250 L
=0.400 M

e) 5.00 gram of glucose (C 6 H 12 O 6 ) diluted to 250 mL
w
% =
v
mass of solute (g)
5.00 g
x 100 %w/v glucose = x 100 = 2.00 %
volume of solution (ml)
250 ml
Molarity =
mole of solute (g)
volume of solution (L)
Molarity ethanol =
1mole
5.00 g x
180 g
0.250 L
= 0.111 M
2. A solution is prepared by dissolving 12.15 gram of Nickel (II) nitrate in 175 mL of
water (density = 1.00 g/ml). Calculate:
a) the % w/w (mass percent)
mass of solution = 12.15 g Nickel + 175 g Nickel = 187.15 g solution.
w
% =
w
mass of solute (g)
mass of solution (ml)
12.15 g
x 100 %w/w Nickel = x 100 = 6.49 %
187.15 g
b) the mole fraction of Ni 2+ ions
Mole Nickel ions =
1mole
12.15 g x = 0.207 mole Ni
58.7g
2+
Mole of H 2 O = 175 g x
1mole
18 g
= 9.72 mole H
2
O
Mole fraction =
2+
mole of Ni
2+
mole of Ni + mole of
H
2
O
Mole fraction of Ni 2+ =
0.207 mole
0.207 mole + 9.72 mole
= 0.0209

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.

5. A solution is prepared by diluting 225 mL of 0.1885 M aluminum sulfate,
Al 2 (SO 4 ) 3 solution with water to a final volume of 1.450L.
Calculate:
a) the number of mole Al 2 (SO 4 ) 3 before dilution
mole of Al 2 (SO 4 ) 3 = 0.225 L x
0.1885 mole
1L
= 0.0424 mole.
b) the molarities of Al 2 (SO 4 ) 3 , Al 3+ ions, SO 4 2- ions in the dilution solution
Molarity of the diluted solution =
0.0424 mole
1.450 L
= 0.0292 M Al 2 (SO 4 ) 3
Molarity of Al 3+ ions = 2 mole Al 3+ :1 mole Al 2 (SO 4 ) 3 = 00585 M Al 3+
Molarity of SO 4 2- ions = 3 mole SO 4
2-
: 1 mole Al 2 (SO 4 ) 3 = 0.127 M
6. A bottle of phosphoric acid is labeled “85.0% H 3 PO 4 by mass” Calculate the
molarity and mole fraction of phosphoric acid in solution.
Mass of H 3 PO 4 = 85 gram
Mole of H 3 PO 4 = 85 gram x
1mole
98 gram
= 0.867 mol
Mass of solution = 100 gram
Assume that density of the solution is 1g/ml
Volume of solution = 100 mL, and grams of H 2 O = 100-85 =15 grams
Mole H 2 O = 15 grams x
1mole
18 gram
= 0.833 mole
Molarity of H 3 PO 4 =
0.867 mole H
3PO
4
0.100 L
= 8.67 M
Mole fraction of H 3 PO 4 = Mole fraction =
mole of H
mole of
3
PO
4
H
3
PO
4
+ mole of
H
2
O
0.867
0.867 mole + 0.833 mole
= 51.0%