1. What precipitate forms when aqueous lead(II) nitrate reacts with aqueous sodium sulfate?
PbSO4
2. A flask contains 45.0 mL of 6.7 M Li2CO3 (MM = 73.888g/mol) solution.
a. How many grams of lithium are present?
? g Li2CO3 = 45.0 mL ( 1 L ) (6.7 mol ) ( 73.888g ) = 22 g Li2CO3
( 1000 mL) ( 1 L ) ( 1 mol )
b. There are 5.0 mL removed from the 45.0 mL solution. The 5.0 mL are then diluted to a total
volume of 125 mL. What is the molarity of the new solution?
? mol Li2CO3 = 5.0 mL ( 1 L
) (6.7 mol Li2CO3) = 0.0335 moles,
(1000 mL)
( L)
Note: 125 mL ( 1 L
) = 0.125 L
(1000 mL)
? M Li2CO3 = 0.0335 mol/0.125 L = 0.268 M Li2CO3
3. How many moles of ammonium ions are produced when 1.6 mol of ammonium phosphate,
(NH4)3PO4, is dissolved in water? Using the same information how many total moles of ions are
produced?
(NH4)3PO4  3 NH4+ (aq) + PO4-3(aq)
? mol NH4+ = 1.6 mol (NH4)3PO4 ( 3 mol NH4+
)
(1 mol (NH4)3PO4)
=
4.8 mol NH4+
Note: there are 1.6 mol PO34Total moles of ions:
4.8 mol NH4+
+ 1.6 mol PO43- = 6.4 total moles
4. Copper(II) bromide is used as a wood preservative. What mass of CuBr2 is needed to prepare
750.0 mL of a 1.25 M solution (MM CuBr2 = 223.35 g/mol).
209g CuBr2
5. Sodium phosphate is used in the preparation of a salt solution. Calculate the
molarity of a solution prepared by dissolving 238 g of Na3PO4 in water to produce 275 mL of
solution.
? mol Na3PO4
Note: 275 mL ( 1 L
(1000 mL)
= 238g Na3PO4 X 1 mol Na3PO4 = 1.45 mol Na3PO4
164 g Na3PO4
) = 0.275 L
1.45 mol Na3PO4/0.275 L = 5.28 M Na3PO4
6. Potassium fluoride is used for frosting glass. Calculate the molarity of a solution prepared by
dissolving 78.6 g of KF in water to produce 225 mL of solution.
? mol KF = 78.6g KF X 1mol KF = 1.35 mol KF
58.10 g
Note: 225 mL ( 1 L
) = 0.225 L
(1000 mL)
1.35 mol KF /0.225 L = 6.01 M KF
7. A 20.0 mL solution (d = 1.2 g/mL) containing 15 g of CuSO4 (MM= 159.61 g/mol) is reacted
with a 30.0 mL solution of 2.0 M lithium hydroxide.
a . What is the %(m/m) of the CuSO4 solution before it is mixed with the lithium hydroxide?
Mass solution: 20.0 mL (1.2 g?1mL) = 24 g solution
Mass solute:
15 g CuSO4
%(m/m):
(15g/24g) 100% = 62 %
b. Write a balance reaction for this precipitation reaction.
CuSO4 (aq) + 2 LiOH (aq)  Cu(OH)2 (s) + Li2SO4 (aq)
c. What is the limiting reagent?
15 g (1 mol CuSO4) (1 mol Cu(OH)2)
( 159.61 g ) ( 1 mol CuSO4 )
= 0.094 mol Cu(OH)2
0.030 L (2.0 mol LiOH) (1 mol Cu(OH)2) = 0.030 mol Cu(OH)2 0.030 < 0.094
( 1L )
( 2 mol LiOH )
LiOH is the limiting Reagent
d. If 2.2 g of copper (II) hydroxide is produced, what is the percent yield?
0.030 mol Cu(OH)2 (97.56 g) = 2.9 g Theoretical Yield
( 1 mol )
Percent yield: (2.2 g /2.9) 100% = 75%
Colligative properties related to change in boiling point and freezing point temperature.
1. When 56 grams of sodium chloride is added to 351 grams of water, what will the
melting point be of the resulting solution? Kf (H2O) = 1.86 °C/m.
2. When 39.5 g of water is used to make a 0.222 m aqueous solution of sucrose, C12H22O11, what is
the mass of sucrose is in the solution? kb= 0.512 °C/m? What is the boiling point of the solution?
3. Find the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol
(C10H18O), a fragrant substance found in the leaves of eucalyptus tree. Kf for CHCl3 is 4.68 °C/m
and the freezing point of pure CHCl3 is -63.5 °C. HINT: CHCl3 is the solvent.