Assignment Solutions CHEM 110- Fall 2011 1. Express the following ...
Assignment Solutions CHEM 110- Fall 2011 1. Express the following ...
Assignment Solutions CHEM 110- Fall 2011 1. Express the following ...
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<strong>Assignment</strong> <strong>Solutions</strong> <strong>CHEM</strong> <strong>110</strong>- <strong>Fall</strong> <strong>2011</strong><br />
<strong>1.</strong> <strong>Express</strong> <strong>the</strong> <strong>following</strong> in %w/v and molarity.<br />
a) 5.62 g of NaHCO 3 dissolved in water to make a 250 mL solution<br />
w<br />
% =<br />
v<br />
mass of solute (g)<br />
5.62 g<br />
x 100 %w/v NaHCO 3 = x 100 = 2.25 %<br />
volume of solution (ml)<br />
250 ml<br />
Molarity =<br />
mole of solute (g)<br />
volume of solution (L)<br />
1mole<br />
5.62 g x<br />
84 g<br />
Molarity NaHCO 3 =<br />
0.250 L<br />
=0.268 M<br />
b) A 500.0 mL solution containing 184.6 g K 2 Cr 2 O 7<br />
w<br />
% =<br />
v<br />
mass of solute (g)<br />
184.6 g<br />
x 100 %w/v K 2 Cr 2 O 7 = x 100 = 36.9 %<br />
volume of solution (ml)<br />
500 ml<br />
Molarity =<br />
mole of solute (g)<br />
volume of solution (L)<br />
1mole<br />
184.6 g x<br />
294 g<br />
Molarity K 2 Cr 2 O 7 =<br />
0.500 L<br />
=<strong>1.</strong>26 M<br />
c) A 250 mL solution containing 17.6 grams of ascorbic acid<br />
w<br />
% =<br />
v<br />
mass of solute (g)<br />
17.6 g<br />
x 100 %w/v ascorbic acid = x 100 = 7.04 %<br />
volume of solution (ml)<br />
250 ml<br />
Molarity =<br />
mole of solute (g)<br />
volume of solution (L)<br />
Molarity ascorbic acid =<br />
1mole<br />
17.6 g x<br />
176 g<br />
0.250 L<br />
=0.400 M<br />
d) 4.6 gram of ethanol diluted to 250 mL.<br />
w<br />
% =<br />
v<br />
mass of solute (g)<br />
4.6 g<br />
x 100 %w/v ethanol = x 100 = <strong>1.</strong>84 %<br />
volume of solution (ml)<br />
250 ml<br />
Molarity =<br />
mole of solute (g)<br />
volume of solution (L)<br />
Molarity ethanol =<br />
1mole<br />
4.6 g x<br />
46 g<br />
0.250 L<br />
=0.400 M
e) 5.00 gram of glucose (C 6 H 12 O 6 ) diluted to 250 mL<br />
w<br />
% =<br />
v<br />
mass of solute (g)<br />
5.00 g<br />
x 100 %w/v glucose = x 100 = 2.00 %<br />
volume of solution (ml)<br />
250 ml<br />
Molarity =<br />
mole of solute (g)<br />
volume of solution (L)<br />
Molarity ethanol =<br />
1mole<br />
5.00 g x<br />
180 g<br />
0.250 L<br />
= 0.111 M<br />
2. A solution is prepared by dissolving 12.15 gram of Nickel (II) nitrate in 175 mL of<br />
water (density = <strong>1.</strong>00 g/ml). Calculate:<br />
a) <strong>the</strong> % w/w (mass percent)<br />
mass of solution = 12.15 g Nickel + 175 g Nickel = 187.15 g solution.<br />
w<br />
% =<br />
w<br />
mass of solute (g)<br />
mass of solution (ml)<br />
12.15 g<br />
x 100 %w/w Nickel = x 100 = 6.49 %<br />
187.15 g<br />
b) <strong>the</strong> mole fraction of Ni 2+ ions<br />
Mole Nickel ions =<br />
1mole<br />
12.15 g x = 0.207 mole Ni<br />
58.7g<br />
2+<br />
Mole of H 2 O = 175 g x<br />
1mole<br />
18 g<br />
= 9.72 mole H<br />
2<br />
O<br />
Mole fraction =<br />
2+<br />
mole of Ni<br />
2+<br />
mole of Ni + mole of<br />
H<br />
2<br />
O<br />
Mole fraction of Ni 2+ =<br />
0.207 mole<br />
0.207 mole + 9.72 mole<br />
= 0.0209
3. For a solution of acetic acid (CH 3 COOH) to be called “vinegar”, it must contain<br />
5.00% acetic acid by mass (mass percent %w/w). Vinegar is a solution of acetic<br />
acid in water. What is <strong>the</strong> molarity of acetic acid in <strong>the</strong> vinegar? The density of<br />
vinegar is <strong>1.</strong>006 g/ml.<br />
CH 3 COOH mass = 5.00 grams in 100 grams of solution for 5.00% (w/w).<br />
Mole CH 3 COOH = 5.00 g x<br />
1mole<br />
60 grams<br />
= 0.0833 mole CH 3 COOH<br />
Volume of solution = 100 grams of solution x<br />
1ml<br />
<strong>1.</strong>006 g<br />
= 99.4 mL<br />
Molarity =<br />
mole of solute<br />
volume of solution (L)<br />
=<br />
0.0833 mole Acetic acid<br />
0.0994 L<br />
= 0.838 M<br />
4. How would you prepare 465 mL of 0.3550 M potassium dichromate (K 2 Cr 2 O 7 )<br />
solution starting with<br />
a) solid potassium dichromate<br />
Determine mole of K 2 Cr 2 O 7 required = 0.465 L x 0.3550<br />
mole = 0.165 mole<br />
L<br />
Determine mass of K 2 Cr 2 O 7 = 0.165 mole x<br />
gram<br />
2 .94 = 48.5 grams<br />
mole<br />
Measure and dissolve 48.5 gram in a total solution volume of 465 ml.<br />
b) 0.750 M potassium dichromate solution.<br />
M 1 V 1 = M 2 V 2<br />
M 1 = 0.750 M : V 1 = ? M 2 = 0.3550 M; V 2 = 465 mL<br />
(0.750 M)(V 1 ) = (0.3550 M) (465 mL)<br />
V 1 =<br />
(0.3550 M)(465 mL)<br />
0.750 M<br />
= 220.1 mL<br />
Measure 220.1 mL of 0.750 M K 2 Cr 2 O 7 into an appropriate size flask <strong>the</strong>n dilute<br />
with water until <strong>the</strong> total solution volume is 465 mL.
5. A solution is prepared by diluting 225 mL of 0.1885 M aluminum sulfate,<br />
Al 2 (SO 4 ) 3 solution with water to a final volume of <strong>1.</strong>450L.<br />
Calculate:<br />
a) <strong>the</strong> number of mole Al 2 (SO 4 ) 3 before dilution<br />
mole of Al 2 (SO 4 ) 3 = 0.225 L x<br />
0.1885 mole<br />
1L<br />
= 0.0424 mole.<br />
b) <strong>the</strong> molarities of Al 2 (SO 4 ) 3 , Al 3+ ions, SO 4 2- ions in <strong>the</strong> dilution solution<br />
Molarity of <strong>the</strong> diluted solution =<br />
0.0424 mole<br />
<strong>1.</strong>450 L<br />
= 0.0292 M Al 2 (SO 4 ) 3<br />
Molarity of Al 3+ ions = 2 mole Al 3+ :1 mole Al 2 (SO 4 ) 3 = 00585 M Al 3+<br />
Molarity of SO 4 2- ions = 3 mole SO 4<br />
2-<br />
: 1 mole Al 2 (SO 4 ) 3 = 0.127 M<br />
6. A bottle of phosphoric acid is labeled “85.0% H 3 PO 4 by mass” Calculate <strong>the</strong><br />
molarity and mole fraction of phosphoric acid in solution.<br />
Mass of H 3 PO 4 = 85 gram<br />
Mole of H 3 PO 4 = 85 gram x<br />
1mole<br />
98 gram<br />
= 0.867 mol<br />
Mass of solution = 100 gram<br />
Assume that density of <strong>the</strong> solution is 1g/ml<br />
Volume of solution = 100 mL, and grams of H 2 O = 100-85 =15 grams<br />
Mole H 2 O = 15 grams x<br />
1mole<br />
18 gram<br />
= 0.833 mole<br />
Molarity of H 3 PO 4 =<br />
0.867 mole H<br />
3PO<br />
4<br />
0.100 L<br />
= 8.67 M<br />
Mole fraction of H 3 PO 4 = Mole fraction =<br />
mole of H<br />
mole of<br />
3<br />
PO<br />
4<br />
H<br />
3<br />
PO<br />
4<br />
+ mole of<br />
H<br />
2<br />
O<br />
0.867<br />
0.867 mole + 0.833 mole<br />
= 5<strong>1.</strong>0%