Chem 32
Solutions to Section 5.8 – 5.10 Homework Problems
.
5.140 The concentration of the solution is 2.6% (w/v). For any isotonic solution, the total
solute concentration is around 0.28 M, so an isotonic solution of glycerol contains 0.28 moles of
glycerol in one liter of solution. To calculate the percentage, convert moles into grams and liters
into milliliters.
0.28 mol ×
92.094 g
= 25.78632 g of C 3H 8O 3
1 mol
1 L = 1000 mL
€ percentage formula to calculate the percent concentration:
Finally, use the
25.78632 g
× 100 = 2.6% (w/v)
1000 mL
5.142 The molarity of isotonic KI is roughly 0.14 M. The total solute concentration must be
0.28 M. Since KI dissociates into K+ and I– ions in solution, we get two moles of ions for every
€
mole of KI we use. Therefore,
we only need 0.28 ÷ 2 = 0.14 moles of KI in each liter of
solution.
5.144 The total molarity of the solution is 0.13 M. You get the total molarity by adding up the
molarities of all of the solutes, being sure to account for any electrolytes that dissociate.
KNO3 is an ionic compound (because K is a metal), so it is a strong electrolyte. This
compound dissociates into K+ and NO3– ions when it dissolves. Therefore, when we dissolve
0.05 moles of KNO3, we make 0.05 moles of K+ and 0.05 moles of NO3– ions. Glucose is a
nonelectrolyte, so the solution contains 0.03 moles of glucose. The total molarity is the sum of
these: 0.05 M + 0.05 M + 0.03 M = 0.13 M
5.150 When you dissolve 0.1 moles of the compound, you form 0.1 x 3 = 0.3 moles of ions.
Therefore, a 0.1 M solution of such a compound has a total ion concentration of 0.3 M,
reasonably close to the isotonic concentration (0.28 M).
5.152 a) Osmosis does not occur, because the total concentration of solutes is 0.4 M on both
sides of the membrane. Remember that NaCl dissociates into two ions. Solution C contains 0.1
mol/L of Na+ ions, 0.1 mol/L of Cl– ions, and 0.2 mol/L of glucose. Solution D contains 0.15
mol/L of Na+ ions, 0.15 mol/L of Cl– ions, and 0.1 mol/L of glucose.
b) Cl– dialyzes from right to left (from solution D to solution C), because there is a higher
concentration of Cl– in solution D.
c) Glucose dialyzes from left to right (from solution C to solution D), because there is a
higher concentration of glucose in solution C.
5.154 a) Initially, osmosis does not occur, because the total solute concentration is 0.1 M on
both sides of the membrane. (The NaCl dissociates into Na+ and Cl– ions, so solution G contains
0.05 M Na+ and 0.05 M Cl–, making a total concentration of 0.1 M.)
b) The problem tells us that ethanol can pass through the membrane, but the sodium and
chloride ions cannot. Therefore, ethanol dialyzes, moving from right to left.
c) After a few minutes, osmosis occurs, with the water moving from right to left. As
ethanol moves from solution H into solution G, the total concentration of solutes in solution G
becomes higher than that in solution H. Water always moves from the solution with the lower
solute concentration to the solution with the higher solute concentration.
5.156 a) This solution is roughly isotonic, because CaCl2 dissociates into three ions, so the total
solute concentration is 0.09 M x 3 = 0.27 M.
b) This solution is hypotonic. NaCl dissociates into two ions, so the total solute
concentration is 0.09 M x 2 = 0.18 M, which is well below the isotonic value.
c) This solution is hypotonic. Glucose does not dissociate, so the total solute
concentration in this solution is 0.09 M.
5.158 a) The cell shrivels up (crenates), because the concentration of NaCl in this solution is
higher than the isotonic value (0.9%).
b) The cell does not change, because 0.9% NaCl is isotonic.
c) The cell swells and bursts (hemolyzes), because the concentration of NaCl in this
solution is lower than the isotonic value (0.9%).
5.160 One mole of PO43– equals 3 equivalents, so we have 0.1 x 3 = 0.3 equivalents (0.3 Eq).
There are 1000 milliequivalents in one equivalent, so we have 0.3 x 1000 = 300 milliequivalents
(300 mEq).
5.162 a) 0.25 moles. One mole of Fe3+ equals three equivalents:
1 mol
0.75 Eq ×
= 0.25 mol
3 Eq
b) 0.008 moles. 24 mEq equals 0.024 Eq, which equals 0.008 moles.
5.164 The solution contains 0.00999 Eq, which equals 9.99 mEq. (The calculator answers are
€
0.00999375 Eq and 9.99375
mEq.) You must convert milligrams into grams, then grams into
moles, then moles into equivalents. You can do this step by step, or in a single sequence of
conversion factors. Here are the individual steps:
480 mg ×
0.48 g ×
1g
= 0.48 g of SO 421000 mg
1 mol
= 0.004996877 moles of SO 4296.06 g
0.004996877 mol ×
2 Eq
= 0.00999375 Eq of SO 421 mol
0.00999375 Eq ×
1000 mEq
= 9.99375 mEq
1 Eq
5.166 a) 2500 mEq. Remember that there are 1000 mEq in one Eq.
b) 0.833 moles. (The calculator answer is 0.833333333 moles.) One mole of Cr3+ equals
three Eq, €
so you must divide the number of moles by 3. The conversion factor setup for this is:
1 mol
2.5 Eq ×
= 0.833333333 mol
3 Eq
d) 43 g. (The calculator answer is 43.33333333 g.) You should calculate this by
converting moles into grams, using the atomic weight of Cr (1 mole equals 52.00 g). Here is the
setup:
€
52.00 g
0.833333333 mol ×
= 43.33333333 g
1 mol
e) 43,000 mg. There are 1000 mg in one gram.
5.168 There are 0.676 g of Cl– in this solution. (The calculator answer is 0.67620875 g.) Here
are the steps:€
Use the concentration to convert the volume (175 mL = 0.175 L) into the number of mEq of Cl–:
109 mEq
0.175 L ×
= 19.075 mEq of Cl1L
Convert mEq into Eq:
1 Eq
19.075 mEq ×
= 0.019075 Eq of Cl−
1000 mEq
€
Convert Eq into moles. Since one mole of Cl– equals one equivalent, there is no calculation to
do: we have 0.019075 moles of Cl–.
€ into grams:
Convert moles
0.019075 mol ×
35.45 g
= 0.676 g of Cl−
1 mol
5.170 a) The molar concentration is 0.0021 M (the calculator answer is 0.00208381 M). First
change 35 mg into grams, and then into moles. 35 milligrams is the same as 0.035 grams, and
€
then…
1 mol
= 0.000364667 moles of HPO 24
95.978 g
Now divide the moles by the volume (in liters) to get the molarity:
0.000364667 mol
= 0.00208381 mol/L
0.175 L
€
0.035 g ×
b) The concentration of this solution is 4.2 mEq/L (the calculator answer is 4.16762
mEq/L). From part a, we know that a liter of solution contains 0.00208381 moles of HPO42-.
The charge on this€ion is –2, so 1 mole of HPO42– equals 2 equivalents:
2 Eq
0.00208381 mol ×
= 0.00416762 Eq
1 mol
Now change this into milliequivalents, moving the decimal point three places to the right. This
gives us 4.16762 mEq in each liter of solution, so the concentration is 4.16762 mEq/L.
c) This€part was assigned by accident – you are not responsible for learning how to
calculate concentrations in mg/dL. (The correct value is 20 mg/dL, if you’re curious.)
d) This part was also assigned by accident – you are not responsible for learning how to
calculate concentrations in ppm. (The correct value is 200 ppm, if you’re curious.)
5.176 You must use 160 mL of the 10% NaOH solution. Use the dilution formula to get the
starting volume:
10% x V1 = 4% x 400 mL
Solving this gives V1 = 160 mL. (To solve, divide both sides by 10%.)
5.178 The concentration of lead is 51 ppb (the calculator answer is 50.6122449 ppb). Use the
dilution formula to get the final concentration:
62 ppb x 80 mL = C2 x 98 mL
Solving this gives C2 = 51 ppb. (To solve, divide both sides by 98 mL.)
5.184 The final volume is roughly 700 mL (the calculator answer is 714.2857143 mL), and you
must add roughly 600 mL of water (the calculator answer is 614.2857143 mL). Use the dilution
formula to find the final volume:
1 M x 100 mL = 0.14 M x V2
Solving this gives V2 = 700 mL. (To solve, divide both sides by 0.14 M.) We started with 100
mL of solution, so we must add 700 – 100 = 600 mL of water.
5.186 The final concentration is 0.2 M. The key here is to realize that the final volume of the
solution is 125 mL (100 mL + 25 mL). Use the dilution formula to get the final concentration of
the solution:
1 M x 25 mL = C2 x 125 mL
Solving this gives C2 = 0.2 M. (To solve, divide both sides by 125 mL.)