Name__________________________________________Date______________________________Period__
Review for Solutions Test
I.
Molarity:
1) What is the molarity of a solution that contains 0.25 moles of magnesium oxalate in 3.65 mL of solution?
2)
3)
4)
5)
6)
7)
8)
II.
Molarity = moles solute/ L solution=
0.25 moles/0.00365 L= 68.5 M
68 M (sf)
What is the molarity of a solution that contains 4.53 moles of lithium nitrate, LiNO3 in 2.85 liters of solution?
Molarity = moles solute/ L solution=
4.53 moles/2.85 L=
1.59 M
A flask contains 85.5 g C12H22O11 (sucrose) in 1.00 liter of solution. What is the molarity?
Moles sucrose= 85.5 g/ 342 g molar mass=
0.25 moles sucrose
Molarity = moles solute/ L solution=
0.25 moles/1.00 L=
0.250 M
Calculate the molarity if a flask contains 21.54 g of potassium sulfate in 2250 ml of solution.
Moles K2SO4= 21.54 g/ 174.2 g molar mass=
0.124 moles K2SO4
Molarity = moles solute/ L solution=
0.124 moles/2.25 L=
0.0551 M
What volume of a 0.02M ammonium nitrate (NH4NO3) solution would contain 3.0 g of ammonium nitrate?
Molarity = moles solute/ L solution
Moles NH4NO3= 3.0 g/ 80.0 g molar mass= 0.0375 moles NH4NO3
0.02 M= 0.0375 moles/X
X= 1.9 L
2 L (sf)
How many grams of hydrochloric acid HCl would be in 5.00 L of a 1.2 M solution?
Molarity = moles solute/ L solution
1.2 M= X/ 5.00 L
X= 6 moles HCl* 36.5g molar mass= 219 g HCl
How many moles of magnesium oxalate (MgC2O4) are in 31.00 ml of a 5.0 M magnesium oxalate solution?
Molarity = moles solute/ L solution
5.0 M= X/0.03100 L
X=0.16 moles MgC2O4
A rabbit weighs out 250 grams of copper (II) sulfate into a flaks and adds enough water to make a
2.0M solution. What is the volume of that solution?
Molarity = moles solute/ L solution
2.0 M= 1.567/X
X=0.78 L
# moles= 250 g CuSO4/159.5 g molar mass=
1.567 moles CuSO4
1)
If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted
solution be?
250*0.75= 295*M
187.5= 295*M
0.64 M
2)
If water is added to 175 mL of a 0.45 M KOH solution until the volume is 250 mL, what will the molarity
of the diluted solution be?
175*0.45= 250*M
78.75= 250*M
0.32 M
3)
How much 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl?
450*9.0= 0.075*V
4050= 0.075*V
54 L
4)
If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the
solution is 275 mL, what will the molarity of the solution be?
550*3.50= 275*M
1925= 275*M
7.0 M
Molality:
1) What is the molality of a solution made by adding 2.00 moles of sodium chloride (NaCl) to 1.00 Kg of water?
Molality= moles of solute/ kg of solvent
molality= 2.00 moles/1.00 kg= 2.00 m
2) What is the molality of a solution that contains 0.52 g of glucose (C6H12O6) in 0.150Kg of water?
Molality= moles of solute/ kg of solvent
molality= 0.0029 moles/ 0.150 kg=
0.019 m
Moles of glucose= 0.52 g glucose/ 180 g molar mass=
0.0029 moles
3) How many Kg of solvent is mixed with 3.0 moles of urea to make a 1.2m solution?
Molality= moles of solute/ kg of solvent
1.2 m= 3.0 moles/X=
x= 2.5 kg
4) How much water is needed to make a 0.50 m solution from 3.2 g of NaCl?
Molality= moles of solute/ kg of solvent
0.50 m= 0.055 moles/X X= 0.11 kg
Name__________________________________________Date______________________________Period__
Review for Solutions Test
# moles= 3.2g/ 58.5g molar mass= 0.055 moles NaCl
5) What mass of CH3OH is needed to add to 1.20 kg of water to make a 3.00 m solution?
Molality= moles of solute/ kg of solvent
3.00 m= X/ 1.20 kg
X= 3.6 moles solute
3.6 moles* 32.05 g molar mass of CH3OH= 115.38 g CH3OH
1) What is the boiling point of a solution made by dissolving 31 g of NaCl in 559 g of water?
31/58.5=
0.53 mol
0.53/0.559=
0.95 m
ΔTb=m*i*kb
=(0.95)(2)(0.52) =0.988
100.99 °C
2) Calculate the freezing point of an a nonionizing antifreeze solution containing 388g ethylene glycol, C2H6O2 , and
409 g of water.
388/62.0=
6.26 mol
6.26/0.409=
15.3 m
ΔTb=m*i*kb
=(15.3)(1)(1.86) =28.42
-28.42°C
3) Calculate the boiling point of an ionic solution containing 29.7 g Na2SO4 and 84.4 g water.
29.7/142.1=
0.21 mol
0.21/0.0844= 2.49 m
ΔTb=m*i*kb
=(2.49)(3)(0.52) =0.386
103.88 °C
III.
Percent Concentration
1) What is the percent concentration by mass of 5.0 g of NaCl mixed with 80. g of water?
(5.0 g)/ (5.0g+ 80.0g)*100=
5.9 % NaCl
2) What mass of NaOH is found in 40. g of a 10% by mass solution?
10= (x/40)*100 0.1=x/40
X=4 g NaOH
3) What is the percent by volume if 22 mL of CO2 is dissolved in enough water to make 85 mL of solution?
(22 mL)/ (85 mL)*100= 25% CO2
75% H2O
4) What volume of ethylene glycol (in mL) is needed to make 1500 mL of antifreeze, a 60.0 % (by volume) solution?
4 qt = 1 gallon. 1 qt = 946 mL.
60 = (X)/ (1500 mL)*100
0.60= (X)/(1500)
X= 900 mL ethylene glycol
IV.
Electrolytes
Identify each of the following as an electrolyte or nonelectrolyte. Then, identify the number of ions each compound
breaks into in water.
Compound
Electrolyte?
# of Ions?
Ca(NO3)2
Yes
3
HCN
Yes
2
Ca(OH)2
Yes
3
Li2SO3
Yes
3
H2SO4
Yes
3
C6H12O6
no
1