Hints for Assignment 1

 
 

1, 2. Hint. Use the periodic table on the inside front cover of your text. Be able to write the symbol if you are given the name (and vice versa) for the first 30 elements plus bromine, silver, tungsten, lead, xenon, arsenic, gold, mercury, iodine, and krypton.
 
 

3. Hint. For a worked example, see "Rule 2" inGuidelines for determining significant figures, p. 47 in Kotz and Treichel (4th ed.)
 
 

4. Hint. For a worked example, see "Rule 3" inGuidelines for determining significant figures, p. 47 in Kotz and Treichel (4th ed.). Remember that the volume of a box is

Volume = length x width x height





5. Sample problem. If PV = [gR(T+273.15)]/M, solve for M when P = 0.1284, V = 0.0879, R = 0.08206, g = 0.1063, and T = 75.

Method: keep track of significant figures with each operation, but don't round until the final result. Your calculator has all those intermediate figures so don't repunch them, just store them and use them without rounding. This makes certain that you don't round too soon.

PV = (0.1284)(0.0879) = 0.011286... (not rounded, 3 s.f. when it is)

T+273.15 = 75+273.15 = 348.15 (not rounded, should be 3 s.f.)

Now: PV has 3 s.f.
g has 4 s.f.
R has 4 s.f.
T + 273.15 has 3 s.f.

Therefore the final answer for M should have 3 s.f.

0.011286... = [0.1063 x 0.08206 x 348.15]/M

M = 269 (rounded from 269.08)

6. Hint. Remember that density is the property that relates mass to volume:
density = mass / volume
so you can get any one of these if you know the value of the other two. Also remember that
volume of the metal sheet = length x width x thickness
So, you can get the volume of the metal sheet given the mass and density, and then get the thickness since the length and width are known. Watch your units in this problem!
 
 

7. Sample problem. Suppose you make a solution that contains 78 g of spurious chloride and 1075 g water.

(a) What is the mass % of the spurious chloride?

(b) If 32.98 mL of your solution weighs 37.421 g, what is the density (in g/cc)?

(a) The mass % of one component is the mass of that component divided by the total mass of all components, multiplied by 100. Thus:
mass % spurious chloride = 100 x [78 g /(78 g + 1075 g)] = 6.8% (rounded to 2 s.f.)
(b) 1 mL = 1 cc = 1 cm3. Thus density = mass/volume = 37.421 g/32.98 mL = 1.135 g/cc
 
 

8. Hint. See Examples 1.1, p. 30, and 1.5, p. 41 for parts (a) and (b). For part (c), remember that density = mass/volume and watch your units.
 
 

9. Sample problem. The density of a solution of spurious chloride is 1.216 g/cc, and it is 10.1% by mass. What volume of the solution (in milliliters) do you need to supply 42.0 g of spurious chloride?

A solution that is 10.1% by mass of spurious chloride contains 10.1 g of spurious chloride in 100 g of solution. Thus, to get 42.0 g spurious chloride,

(42.0 g spurious chloride) x (100 g solution/10.1 g spurious chloride)

= 416 g of solution is needed.
Volume required = 416 g solution x 1 cc solution/1.216 g solution
= 342 mL
10. Hints. See Figure 1.2 in Kotz and Treichel (p. 20) for particulate representations of the gaseous , liquid, and solid states of matter. Note that the gaseous and liquid phases are shown for bromine.

Read the caption to Figure 1.2 carefully. It explains how the particulate representation relates to the description of a solid, liquid, or gas.
 


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Last revised: August 20 2000 by M. F. Richardson
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