Problems 11-25

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**Problem #11:** A cylindrical glass tube of length 27.75 cm and the radius 2.00 cm is filled with argon gas. The empty tube weighs 188.250 g. and the tube filled with argon weighs 188.870 g. Use the data to calculate the density of argon gas.

**Solution:**

1) Volume of a cylinder:

V = πr^{2}hV = (3.14159) (2.00 cm)

^{2}(27.75 cm)V = 348.7165 cm

^{3}

2) Mass:

188.870 g - 188.250 g = 0.620 g

3) Density:

0.620 g / 0.3487165 L = 1.78 g/LNote that 348.7165 cm

^{3}became 0.3487165 liters. Gas density is typically measured in g/L as opposed to g/cm^{3}or g/mL. Reminders: 1 cm^{3}= 1 mL and 1000 mL = 1 L.

**Problem #12:** The density of silver is 10.50 g/cm^{3} and the density of benzene is 0.8786 g/cm^{3}. What mass of silver will have the same volume as 15.55 grams of benzene?

**Solution:**

1) Determine the volume of benzene:

mass / density = volume15.55 g / 0.8786 g/cm

^{3}= 17.6986 cm^{3}

2) Determine the mass of silver:

density times volume = mass(10.50 g/cm

^{3}) (17.6986 cm^{3}) = 185.8 g

**Problem #13:** Calculate the mass of copper in grams (density = 8.96 g/cm^{3}) with the same volume as 100.0 grams of gold (density = 19.31 g/cm^{3})

**Solution:**

1) Volume of gold:

100.0 g ÷ 19.31 g/cm^{3}= 5.17866 cm^{3}

2) Mass of copper:

(8.96 g/cm^{3}) (5.17866 cm^{3}) = 46.4 g

3) Setting up the problem in dimensional analysis style:

1 cm ^{3}8.96 g 100.0 g x ––––––– x ––––––– = 46.4 g 19.31 g 1 cm ^{3}

**Problem #14:** Calculate the mass of zinc in grams (density = 7.14 g/cm^{3}) with the same volume as 100.0 grams of aluminum (density = 2.70 g/cm^{3})

**Solution:**

1) Volume of aluminum:

100.0 g ÷ (2.70 g/cm^{3}) = 37.037 cm^{3}

2) Mass of zinc:

(7.14 g/cm^{3}) (37.037 cm^{3}) = 264 g

3) Dimensional analysis:

1 cm ^{3}7.14 g 100.0 g x ––––––– x ––––––– = 264 g 2.70 g 1 cm ^{3}

**Problem #15:** A spherical ball bearing has a radius of 8.50 mm and a mass of 2.315 g. Determine the density of the ball bearing in g/cm^{3}.

**Solution:**

1) Convert mm to cm:

1 cm 8.50 mm x ––––––– = 0.850 cm 10 mm

2) Determine volume of sphere:

V = (4/3)πr^{3}(4/3) (3.14159) (0.850 cm)

^{3}= 2.57 cm^{3}

3) Calculate density:

2.315 g / 2.57 cm^{3}= 0.900 g/cm^{3}

**Problem #16:** 57.0 kg of copper is drawn into a wire with a diameter of 9.50 mm. What is the length of wire in meters? Cu density = 8.96 g/cm^{3}.

**Solution:**

1) Convert kg to grams:

1000 g 57.0 kg x ––––––– = 5.70 x 10 ^{4}g1 kg

2) Determine volume of the copper wire:

(5.70 x 10^{4}g) ÷ (8.96 g/cm^{3}) = 6361.607 cm^{3}

3) Convert mm to cm:

1 cm 9.50 mm x ––––––– = 0.950 cm 10 mm

3) Determine length of wire:

V = πr^{2}h6361.607 cm

^{3}= (3.14159) (0.475 cm)^{2}hh = 8974.91 cm

To three sig figs and in meters, 89.7 m

**Problem #17:** In the United States, 'copper' pennies made since 1983 actually contain very little copper. If a penny contains 93.975% of its total volume zinc and 6.025% of its total volume copper, what is its apparent density? (density of Cu = 8.96 g/cm^{3}; density of Zn = 7.14 g/cm^{3}.)

**Solution:**

1) Assume the penny occupies 1.00 cm^{3}. This means:

copper occupies 0.06025 cm^{3}and zinc occupies 0.93975 cm^{3}.

2) Calculate mass of copper:

(0.06025 cm^{3}) (8.96 g/cm^{3}) = 0.53984 g

3) Calculate mass of zinc:

(0.93975 cm^{3}) (7.14 g/cm^{3}) = 6.709815 g

4) Determine apparent density:

0.53984 g + 6.709815 g = 7.249655 gsince this mass is in 1.00 cm

^{3}, the answer is 7.25 g/cm^{3}

**Problem #18:** Antarctica has an ice sheet covering 1.42 x 10^{18} cm^{2} and averaging 1.61 x 10^{5} cm deep. Calculate the total mass if ice has a density of 0.92 g/cm^{3}.

**Solution:**

1) Calculate volume of ice:

(1.42 x 10^{18}cm^{2}) (1.61 x 10^{5}cm) = 2.2862 x 10^{23}cm^{3}

2) Calculate mass of ice:

(2.2862 x 10^{23}cm^{3}) (0.92 g/cm^{3}) = 2.1 x 10^{23}g

**Problem #19:** Object A is less dense than object B. If both objects are the same mass, what can be said about the volume of A as compared to the volume of B?

**Solution:**

Object A has a larger volume than Object B.

**Problem #20:** An ice cube with a volume of 45.0 mL and a density of 0.900 g/cm^{3} floats in a liquid with a density of 1.36 g/mL. What volume of the cube is submerged in the liquid?

**Solution:**

The solution to this problem involves the concept of buoyancy.

1) Determine the mass of the cube:

(45.0 mL) (0.900 g/cm^{3}) = 40.5 g

2) The cube will float when 40.5 g of liquid is displaced. We need to know what volume of the liquid weighs 40.5 g.

40.5 g) ÷ (1.36 g/mL) = 29.8 mLThis means that 29.8 mL of the cube is submerged (this is the answer to the question), displacing 40.5 g of the liquid. The rest of the cube (45.0 − 29.8) is above the surface of the liquid.

**Problem #21:** Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = 6.304 x 10¯^{3} inches) can be produced from the copper that is in 5.88 pounds of covellite, an ore of copper that is 66% copper by mass? (Hint: treat the wire as a cylinder. The density of copper is 8.96 g cm¯^{3}; one kg weighs 2.2046 lb; 1 inch is 2.54 cm and the volume of a cylinder is πr^{2}h)

**Solution:**

a) Determine pounds of pure copper in 5.88 lbs of covellite:

5.88 lbs x 0.66 = 3.8808 lbs

b) Convert pounds to kilograms:

3.8808 lbs ÷ 2.6046 lbs kg¯^{1}= 1.489979 kg (I'm keeping a few guard digits)1.489979 kg = 1489.979 g

c) Determine volume this amount of copper occupies:

8.96 g cm¯^{3}= 1489.979 g / xx = 166.292 cm

^{3}Note: this is the volume of the cylinder.

d) Convert the diameter in inches to a radius in centimeters:

dia = 6.304 x 10¯^{3}in; radius = 3.152 x 10¯^{3}in(3.152 x 10¯

^{3}inch) (2.54 cm / inch) = 8.00 x 10¯^{3}cm

e) Determine h in the volume of a cylinder:

166.292 cm^{3}= (3.14159) (8.00 x 10¯^{3}cm)^{2}hh = 8.27 x 10

^{5}cm = 8.27 x 10^{3}m

**Problem #22:** A copper ingot has a mass of 2.15 kg. If the copper is drawn into wire whose diameter is 2.27 mm, how many inches of copper wire can be obtained from the ingot?

**Solution:**

a) Determine volume of copper:

8.96 g cm¯^{3}= 2150 g / xx = 239.955 cm

^{3}Note: this is the volume of the wire.

b) Determine h in the volume of a cylinder (i.e., the wire):

239.955 cm^{3}= (3.14159) (0.1135 cm)^{2}hh = 5929.097 cm

Note: 0.1135 cm is the radius

c) Convert cm to inch:

5929.097 cm divided by 2.54 cm/in = 2334.29 inTo three sig figs, the answer is 2330 in

**Problem #23:** If the copper is drawn into wire whose diameter is 8.06 mm, how many feet of copper can be obtained from a 200.0 pound ingot?

**Solution:**

a) Convert pounds to grams:

200.0 lb x (453.59 g/lb) = 9.0718 x 10^{4}g

b) Determine what volume is occupied by this many grams of copper:

8.96 g/cm^{3}= 9.0718 x 10^{4}g divided by xx = 1.01248 x 10

^{4}cm^{3}

c) Determine the height of the cylinder (Volume of a cylinder = πr^{2}h):

1.01248 x 10^{4}cm^{3}= (3.14159) (0.0403 cm)^{2}hh = 1.9844 x 10

^{6}cm

d) Convert cm to inches, then to feet:

1.9822 x 10^{6}cm x (1 in/2.54 cm) = 7.8126 x 10^{5}in7.8126 x 10

^{5}in x (1 ft / 12 in) = 6.51 x 10^{4}ft (to three sig figs)

**Problem #24:** A cube of copper was found to have a mass of 0.630 kg. What are the dimensions of the cube? (The density of copper is 8.96 g/cm^{3}.)

**Solution:**

a) Determine the volume of the cube (note that kg have been converted to g):

8.96 g/cm^{3}= 630. g / volumevolume = 70.3125 cm

^{3}

b) Each side of a cube is equal in length, so take cube root of the volume for length of cube side:

$\sqrt[3]{\mathrm{70.30125\; cm3}}$ = 4.13 cm (to three sig figs)

**Problem #25:** Calculate the volume (in m^{3}) of a 5,020 tonne iceberg. (1 tonne = 1,000 kg, the density of ice = 0.92 g/cm^{3})

**Solution:**

a) Convert tonnes to grams:

5,020 tonne x (1,000 kg / tonne) = 5.02 x 10^{6}kg5.02 x 10

^{6}kg x (1000 g / kg) = 5.02 x 10^{9}g

b) Determine volume in cubic centimeters:

0.92 g/cm^{3}= 5.02 x 10^{9}g / volumevolume = 5.4565 x 10

^{9}cm^{3}

c) Convert cubic centimeters to cubic meters:

5.4565 x 10^{9}cm^{3}x (1 m^{3}/ 10^{6}cm^{3}) = 5.46 x 10^{3}m^{3}(rounded to 3 significant figures)

**Bonus Problem:** A room contains 11.5 kg of argon. The air in the room consists of 0.225% argon. The density of air in the room is 1.70 g/L. What is the volume of the room in m^{3}?

Solution #1:

a) 11.5 kg is 0.225% of the total mass in the room. Determine total mass (in grams) of air in room:

11.5 kg / 0.00225 = 5111 kg5111 kg x (1000 g/kg) = 5.111 x 10

^{6}g

b) Using density, determine volume of room in liters:

5.111 x 10^{6}g ÷ 1.70 g/L = 3.00 x 10^{6}L

c) Convert liters to cubic meters:

3.00 x 10^{6}L x (1 m^{3}/ 1000 L) = 3.00 x 10^{3}m^{3}

Solution #2:

a) Determine the liters that 11.5 kg of argon occupies:

11.5 kg x (1000 g/kg) = 11,500 g11,500 g / 1.70 g/L = 6.7647 x 10

^{3}L

b) The volume the Ar occupies represents 0.225% of the entire room. Determine the total volume of the room:

6.7647 x 10^{3}L / 0.00225 = 3.00 x 10^{6}L

c) Convert liters to cubic meters:

See step (c) in solution #1.

Fifteen Examples | Probs #1-10 | Probs #26-50 | All the examples & problems, no solutions | Significant Figures Menu |