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

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%