Chemistry 2 AP/AC
Unit 2 HW 1
Name_________________________
1) Classify each of the following as an electrolyte (E) or nonelectrolyte (NE).
(a) HBr _____
(b) C12H22O11 _____
(c) KBr _____
(d) KOH _____
(e) CH3CH2OH (ethanol) _____
(f) CaCl2 _____
2) Calculate the mass of KI in grams required to prepare 500.0 mL of a 2.80 M solution.
3) How many mL is required to provide 2.14 g of sodium chloride if the solution of
sodium chloride is 0.270 M?
4) What is the molarity of a solution that has 29.0 g of ethanol (C2H5OH) dissolved in
545 mL of solution?
5) Suppose you mix 35.2 mL of a 1.66 M C6H12O6 solution with 16.7 mL of a 0.892 M
C6H12O6 solution. What would be the molarity of the final C6H12O6 solution?
6) What is the concentration of all ions present in a solution made by combining 20.0 mL
of 1.00 M NaCl with 30.0 mL of 1.20 M CaCl2 with 50.0 mL of 1.60 M AlCl3?
Assume that all of these substances ionize completely in solution.
7) Describe how to prepare 1.00 L of 0.646 M HCl solution, starting with 12.0 M HCl.
8) Water is added to 25.0 mL of a 0.855 M KNO3 solution until the volume of the
solution is exactly 500.0 mL. What is the concentration of the final solution?
9) You have 505 mL of a 0.125 M HCl solution and you want to dilute it to exactly
0.100 M. How much water will be needed?
10) What does it mean when we say volumes are “additive”? When will volumes be
additive?
11) Page 181, #25 (Ammonium Sulfate, of course is (NH4)2SO4)
12) page 181, #28
KEY to Unit 2 HW 1
1) Ionic compounds and acids are electrolytes:
E
NE
E
E
NE
E
2) 0.5000 L (2.80 mol/L) (166.00 g KI / mol KI) = 232 g KI
3) 2.14 g NaCl · 1 mol NaCl
58.44 g NaCl
4)
[C2H5OH] =
·
1L
· 10000 mL = 136 mL
0.270 mol NaCl
1L
29.0 g · 1 mol C2H5OH
58.44 g___
0.545 L
= 1.15 M
5) Total Volume = 35.2 mL + 16.7 mL = 51.9 mL
[C6H12O6] = (0.0352 L) (1.66 mol/L) + (0.0167 L) (0.892 mol/L) = 1.41 M
0.0519 L
6) Total Volume = 20.0 mL + 30.0 mL + 50.0 mL = 100.0 mL
[Na+] = (0.0200 L)(1.00 mol/L) = 0.200 M
0.1000 L
[Cl-] = (0.0200 L)(1.00 mol/L) + (0.0300 L)(2.40 mol/L) + (0.0500 L)(4.80 mol/L)
0.1000 L
= 3.32 M
[Ca+2] = (0.0300 L)(1.20 mol/L) = 0.360 M
0.1000 L
[Al+3] = (0.0500 L)(1.60 mol/L) = 0.800 M
0.1000 L
7) M1V1 = M2V2
(12.0 M)(V1) = (0.646 M)(1.00 L)
V1 = 0.0538 L = 53.8 mL of concentrated HCl needs to be used
VH2O ≈ 1000 mL – 53.8 mL ≈ 946 mL
So, you would need to add 53.8 mL of acid to a little less than 946 mL H2O, then
dilute with water up to 1.00 L.
8) M1V1 = M2V2
(0.855 M)(25.0 mL) = (M2)(500.0 mL)
M2 = 0.0428 M
9) M1V1 = M2V2
(0.125 M)(505 mL) = (0.100 M)(V2)
V2 = 631 mL = total final volume
VH2O = 631 mL – 505 mL = 126 mL H2O (assuming volumes are additive)
10) This statement means that the total volume of a combination of solutions is equal to
the algebraic sum of the individual volumes of the solutions. Volumes are additive
when the particles are similar in size and attractive forces.
11) [(NH4)2SO4] = 10.8 g · 1 mol (NH4)2SO4
132.17 g
1.000 L
M1V1 = M2V2
= 0.817 M
(0.817 M)(10.00 mL) = (M2)(60.00 mL)
M2 = diluted [(NH4)2SO4] = 0.136 M
[NH4+] = (2)(0.136 M) = 0.272 M [SO4-2] = 0.136 M
12) [Mn+2] = 1.584 g · 1 mol Mn+2
54.94 g
1.000 L
= 0.02883 M
Solution A: M1V1 = M2V2
(0.02883 M)(50.00 mL) = (M2)(1000.0 mL)
M2 = 0.001442 M
Solution B: M1V1 = M2V2
(0.001442 M)(10.00 mL) = (M2)(250.0 mL)
M2 = 5.768 X 10-5 M
Solution C: M1V1 = M2V2
(5.768 X 10-5 M)(10.00 mL) = (M2)(500.0 mL)
M2 = 1.154 X 10-6 M