CHEM 150
Assignment #4
(Due Thursday, Nov 21st, in class)
TOTAL = 28 points
1) How many grams of water must be added to 10.0 g of Ca(NO3)2 in order to prepare a 12.0
%(m/m) solution?
( )
(
)
(
)
Mass of solution = mass of solute + mass of solvent, so mass of solvent = 83.333…g – 10.0g = 73.3 g
H2O. 1 for solution mass, 1 for solvent mass. -0.5 if sig figs wrong.
2) After all of the water has been evaporated from a 25.0 mL solution of NaCl, 2.00 g of solid NaCl
remains. Calculate the original concentration of the NaCl solution in the following units:
a. %(m/v)
2.00g NaCl in 25.0 mL of solution:
( )
( )
(
(
)
)
1 point.
b. molarity
molar mass of NaCl = 58.44 g/mol.
(
)(
)
1 for molar mass, 1 for mol NaCl, 1 for concentration. -0.5 for wrong sig figs for a or b.
3) Indicate whether the osmotic pressure of 0.1 M NaCl will be less than, the same as, or greater
than the osmotic pressure of each of the following solutions:
a. 0.1 M NaBr
The osmotic pressure of 0.1M NaBr is proportional to its osmolar concentration:
Osmolarity = M x i
For NaBr  Na+ + Br-
i=2
Osmolarity = (0.1M)(2) = 0.2M
1 point
For 0.1M NaCl, i = 2 and osmolarity = (0.1M)(2) = 0.2M 1 point
So the osmotic pressure of 0.1M NaCl would be expected to be the same as the osmotic pressure of
0.1M NaBr. 1 point
b. 0.050 M MgCl2
For 0.050M MgCl2, i = 3: MgCl2  Mg2+ + 2ClOsmolarity = (0.050M)(3) = 0.15M. 1 point Thus, the osmotic pressure of 0.1M NaCl is greater than the
osmotic pressure of 0.050 M MgCl2. 1 point
c. 0.1 M MgCl2
For 0.1M MgCl2, osmolarity = (0.1M)(3) = 0.3M. 1 point Thus, the osmotic pressure of 0.1M NaCl should
be less than the osmotic pressure of 0.1 M MgCl2. 1 point
d. 0.1 M glucose
For glucose (a molecular compound), i = 1. Thus, the osmotic pressure of 0.1 M glucose would be
(0.1M)(1) = 0.1 M. 1 point Thus, the osmotic pressure of 0.1M NaCl would be expected to be greater
than the osmotic pressure of 0.1M glucose. 1 point
4) Look at slide # 45, chapter 8. If 0.25 mol of NaCl increases the temperature of 1 kg of water by
0.26 oC, what is the boiling point of a solution formed by adding 0.25 mol of NaCl to 0.50 kg of
water?
The boiling point increase is proportional to the concentration of the solute. If the concentration of the
solute doubles, then so does the boiling point increase. In comparing a solution that consists of 0.25
mol NaCl in 1 kg H2O with another one that has 0.25 mol NaCl in only 0.50 kg of water, the second
solution has a concentration that is twice as high as the first one (they have the same amount of solute,
but the second one has this amount of solid dissolved in half the volume of solvent.) Since the second
solution has twice as high a concentration, its boiling point increase will be double what is shown for the
first solution on slide #45 of the notes. Its b.p. should thus be 100.51 oC. 2 points
5) Classify each of the following reactions as redox or non-redox and indicate the oxidizing agent in
each case.
a. 2NO + O2  2NO2
Oxidation numbers change as follows: N(+2)  N(+4); O(0 in O2)  O(-2). There is a
change in oxidation numbers for N and O, so it is redox. 1 point Oxidizing agent is O2. 1
point
b. CO2 + H2O  H2CO3
On both sides, oxidation number for O is -2, for H is +1, and for C is +4. Non-redox. 1
point
c. Zn + 2AgNO3  Zn(NO3)2 + 2Ag
Oxidation numbers: Zn(0  +2); Ag (+1  0). Redox 1 point, and AgNO3 is oxidizing
agent. 1 point (really, Ag+ is)
d. HNO3 + NaOH  NaNO3 + H2O
On both sides, oxidation number for H is +1, O is -2, N is +5, and Na is +1. Non-redox. 1
point
6) Determine the oxidation number for Cr (chromium) in each of the following chromiumcontaining species:
a. Cr2O3
Cr: +3 1 point
b. CrO4Cr: +7 1 point
c. Na2Cr2O7
Cr: +6 1 point
d. CrF5
Cr: +5 1 point
e. CrO3
Cr: +6 1 point