ANSWER GUIDE TO HOMEWORK 8e and 8f
MOLECULAR WEIGHTS FOR HOMEWORK 8e:
H2O water
H3PO4 phosphoric acid
C2H6O2 ethylene glycol
H = 1.01 * 2 = 2.02
H = 1.01 * 3 = 3.03
C = 12.01 * 2 = 24.02
O = 16.00
= 16.00
P = 30.97
= 30.97
H = 1.01 * 6 = 6.06
1 mole = 18.02 g
O = 16.00 * 4 = 64.00
O = 16.00 * 2 = 32.00
1 mole = 98.00 g
1 mole = 62.08 g
1.a. Density of water is 1.00 g/ml, so 100.0 ml has a mass of 100.0 ml * (1.000g /ml) = 100.0 g
Total mass is 10.0 g H3PO4 + 100.0 g water = 110.0 g.
Density of the solution is mass/volume = 110. g / 104.0 ml = 1.058 g /ml
1.b. 10.0 g of H3PO4 * (1 mole / 98.00 g) =
.10204 moles
100.0 g of H2O * (1 mole / 18.02 g) =
5.54939 moles
TOTAL
5.65143 moles
Mole fraction is (.10204 / 5.65143) = .0181 (or 1.81%) phosphoric acid and the rest (98.19%) water
1.c. 10.0 g of H3PO4 * (1 mole / 98.00 g) = .10204 moles; Volume of 104.0 ml = .104 L
Molarity = .10204 moles / .104 L = .981 mol /L = .981 M
1d. Molality = moles of solute / kg of solvent = .10204 moles / .1 kg = 1.02 molal
2. Imagine 100.0 g of this solution.
40.0% of the mass = 40.0 g is ethylene glycol and the remaining 60.0 g (or .060 kg) is water.
40.0 g of C2H6O2 * ( 1 mole / 62.08 g) =
.6443 moles
60.0 g of H2O * (1 mole / 18.02 g) =
3.3330 moles
TOTAL MOLES = 3.9773 moles
Density is 1.05 g/ml, so volume = 100.0 g * (1 ml / 1.05 g) = 95.238 ml or .095238 L
(a) Molarity = .6443 moles / .095238 L = 6.779 M
(b) Molality = .6443 moles / .060 kg = 10.74 moles / kg or 10.74 molal
(c) Mole fraction = .6443 moles / 3.9773 moles = .162, or 16.2% of all the moles
3.a.) 2.0 molal KCl in water MW of KCl = 74.55, so use 0.2 moles * 74.55 g/mol = 14.91 g
ANSWER: Measure out 14.91 g of KCl and mix with 100 g of water.
Solution will then have 0.2 moles per 0.1 kg of water, which is 2.0 moles per kg or 2.0 molal
3.b.) Solution is 15% NaOH (by mass) in water, so solution would contain 85% water.
The final volume of 100.0 ml is approximately the same as the volume of water used, so 100 ml
of water are needed, and this would have a mass of 100 g, amounting to 85% of the total.
The total mass is therefore 100 g * (1 / .85) = 117.65 g, of which 15% = .15 * 117.65 = 17.65 g is NaOH
ANSWER: Measure out 17.65 g of NaOH and mix with 100.0 g of water.
3.c.) Solution is 25% NaOH (by mass) in methanol, so solution would contain 75% methanol.
The final volume of 100 ml is approximately the same as the volume of methanol, so 100 ml
3
of methanol are needed. Since the density of methanol (given at the top of the page) is 0.79 g/cm ,
equivalent to 0.79 g/ml, the methanol would have a mass of 100 ml * 0.79 g /ml = 79.0 g and this would
be 75% or 0.75 of the total mass. The total mass is therefore 79.0 g / 0.75 = 105.33 g, of which 25%, or
.25 * 105.33 = 26.33 g is NaOH.
ANSWER: Measure out 26.33 g of NaOH and mix with 79.0 g of methanol for a total mass of 105.33 g,
of which 25% is NaOH and the rest methanol.
CONTINUED ON NEXT PAGE
3.d.) 0.10 mole fraction of glucose (C6H12O6) in water
MW of glucose = 180.156
A 0.10 mole fraction of glucose must contain 9 moles of water ( = 9 mol * 18.02 g/mol = 162.18 g) for
every 1 mole ( = 180.156 g ) of glucose.
The mass of the glucose must be 180.156 / 162.18 = 1.11084 times the mass of the water.
A solution with a volume of 100.0 ml needs 100.0 g of water (with a density of 1.00 g/ml), and therefore a
mass of glucose equal to 1.11084 times this amount, or 111.084 g.
ANSWER: Measure out 111.084 g of glucose (equivalent to 111.084/180.156 = .6166 mole) and mix
with 100.0 g of water (equivalent to 100.0/18.02 = 5.5494 moles) for a total of 6.1660 moles, 10% of
which are moles of glucose. If the total volume of the solution exceeds 100 ml, just measure out 100 ml
and discard the rest.
4. The vapor pressure is reduced in proportion to the mole fraction of the solvent. If a solution is only
95% water (mole fraction of .95), then the vapor pressure is lowered to 95% of what it would be for pure
water at that temperature. Note also that “torr” is another name for “mm of mercury”, a unit of pressure.
+
–
4.a. For every 100 moles of solution, there is 1 mole of NaCl, and two moles of ions (Na and Cl ) after
dissociation, so the mole fraction of the water is reduced to 98%.
Thus, the vapor pressure is reduced to .98 of its value for pure water, or 0.98 *23.8 = 23.3 torr.
4.b. Glucose does not dissociate, so there is 1 mole of glucose in every 100 moles of solution, so the water
has a mole fraction of .99 . The vapor pressure is then 0.99 *23.8 = 23.6 torr.
MOLECULAR WEIGHTS FOR HOMEWORK 8f:
H2O water
C3H8O3 glycerine
C2H6O2 ethylene glycol
H = 1.01 * 2 = 2.02
C = 12.01 * 3 = 36.03
C = 12.01 * 2 = 24.02
O = 16.00
= 16.00
H = 1.01 * 8 = 8.08
H = 1.01 * 6 = 6.06
1 mole = 18.02 g
O = 16.00 * 3 = 48.00
O = 16.00 * 2 = 32.00
1 mole = 92.11 g
1 mole = 62.08 g
1. ∆T = i m Kf
and i = 1 because glycerine is a nonelectrolyte.
o
So -1.50 C = 1 * m * (-1.86oC/molal)
m = 1.50oC / (1.86oC/molal) = .80645 molal = .80645 moles / kg = x / .2 kg
solving for x, we find x = .1613 moles. Multiplying by the MW of glycerine (above),
we find .1613 moles * (92.11 g/mole) = 14.86 g
2. ∆T = i m Kf
and i = 1 because ethylene glycol is a nonelectrolyte.
If we imagine 200.0 g of this solution, then:
50.0% by mass, or 100.0 g , is ethylene glycol, equal to 100.0 g * (1 mol / 62.08 g) = 1.611 moles.
The other 100 g is water, equal to 0.1 kg of solvent.
The molality, m, is 1.611 moles / 0.1 kg = 16.11 moles / kg = 16.11 molal
Boiling point = 100oC + ∆T = 100oC + i m Kf = 100oC + 1 (16.11 molal) (0.51oC/molal) = 108.2oC
Freezing point = 0oC + ∆T = 0oC + i m Kb = 0oC +
1 (16.11 molal) (-1.86oC/molal) = - 30.0oC
Kf = -1.86oC/molal
a.) MgCl2 breaks up into 3 ions, so i = 3
Boiling point = 100oC + ∆T = 100oC + i m Kf =
Freezing point = 0oC + ∆T = 0oC + i m Kb =
b.) FeCl3 breaks up into 4 ions, so i = 4
Boiling point = 100oC + ∆T = 100oC + i m Kf =
Freezing point = 0oC + ∆T = 0oC + i m Kb =
3. ∆T = i m Kf
∆T = i m Kb Kb = 0.51oC/molal
100oC + 3 (0.050 molal) (0.51oC/molal) = 100.08oC
0oC + 3 (0.050 molal) (-1.86oC/molal) = - 0.28oC
100oC + 4 (0.050 molal) (0.51oC/molal) = 100.10oC
0oC + 4 (0.050 molal) (-1.86oC/molal) = - 0.37oC