Problems 1-10

Twenty Examples | Probs #11-25 | Probs #26-50 | All the examples & problems, no solutions | Significant Figures Menu |

**Problem #1:** What mass of lead (density 11.4 g/cm^{3}) would have an identical volume to 25.1 g of mercury (density 13.6 g/cm^{3})?

**Solution #1:**

1) Determine the volume of 25.1 g of mercury:

M D = –––– V

25.1 g 13.6 g/cm ^{3}= –––––– V V = 1.845588 cm

^{3}

2) Determine the mass of 1.845588 cm^{3} of lead:

M D = –––– V

M 11.4 g/cm ^{3}= –––––––––––– 1.845588 cm ^{3}V = 21.0 g

**Solution #2:**

1) This solution depends on the fact that the volumes are equal. Using symbols, we show the equality as follows:

V_{Pb}= V_{Hg}

2) Rearranging the density equation for lead and for mercury, we have this:

M _{Pb}V _{Pb}= –––– D _{Pb}and

M _{Hg}V _{Hg}= –––– D _{Hg}

3) Setting them equal to each other, we have this:

M _{Pb}M _{Hg}–––– = –––– D _{Pb}D _{Hg}

4) Substitute values into the above equation, then solve:

M _{Pb}25.1 –––– = –––– 11.4 13.6 M

_{Pb}= 21.0 g

**Problem #2:** The density of pure aluminum is 2.70 g/cm^{3}. 2.25 grams of pure aluminum is added to a graduated cylinder containing 11.20 mL of water. To what volume mark will the water level rise?

**Solution:**

1) Determine the volume of 2.25 g of aluminum:

Density = mass / volumerearrange to:

Volume = mass / density

V = 2.25 g / 2.70 g/cm

^{3}= 0.83 cm^{3}Note: since 1 mL = 1 cm

^{3}, this is 0.83 mL

2) Add:

11.20 mL + 0.83 mL = 12.03 mL

**Problem #3:** A 1.50 g piece of aluminum was rolled out into a thin sheet measuring 24.0 cm x 30.0 cm.

(a) Calculate the volume (density of Al = 2.70 g cm¯^{3}

(b) Calculate the thickness of the piece of Al in mm

**Solution:**

1.50 g / 2.70 g/cm^{3}= 0.555555 cm^{3}(answer to a)0.555555 cm

^{3}= (x) (24.0 cm) (30.0 cm)x = 0.000772 cm (to 3 sig figs) <--- the thickness of the Al foil in cm.

There are 10 mm in one cm, so divide the cm answer by 10 to get the thickness in mm.

0.0000772 mm (answer to b)

**Problem #4:** 'Copper' pennies actually contain very little copper. If a penny contains 92.276% of its total volume zinc and 7.724% of its total volume copper, what is its density? (d of Cu = 8.96 g/cm^{3}; d of Zn = 7.14 g/cm^{3})

**Solution:**

1) Let's assume a penny is 100. cm^{3} for its volume:

That means 92.276 cm^{3}is Zn and 7.724 cm^{3}is Cu.

2) Let's figure out the mass of Zn and the mass of Cu:

(92.276 cm^{3}) (7.14 g/cm^{3}) = 658.85064 g

(7.724 cm^{3}) (8.96 g/cm^{3}) = 69.20704 g

3) Now, we add the masses and divide by the total volume:

728.05768 g / 100. cm^{3}= 7.28 g/cm^{3}(that's the answer to three sig figs)

**Problem #5:** The element antimony has a density of 6.62 g cm¯^{3}. Calculate the edge length in meters of a cube of antimony whose mass is 1.00 x 10^{5} kg.

**Solution:**

1.00 x 10^{5}kg = 1.00 x 10^{8}g1.00 x 10

^{8}g / 6.62 g cm¯^{3}= 1.51 x 10^{7}cm^{3}$\sqrt[3]{\mathrm{1.51\; x\; 107cm3}}$ = 247 cm = 2.47 m

**Problem #6:** Iron has a density of 7.87 g/cm^{3}. If 71.7 g of iron is added to 63.0 mL of water in a cylinder, to what volume reading will the water rise?

**Solution:**

1) We need to determine the volume of iron:

D = m / VV = m / D

V = 71.7 g / 7.87 g/cm

^{3}V = 9.11 cm

^{3}

2) Since 1 cm^{3} = 1 mL, we have this:

63.0 mL + 9.11 mL = 72.1 mL (rounded off to the proper number of sig figs)

3) You can approach this problem using dimensional analysis (also called the unit-factor method or the factor-label method), where the density is used as a conversion factor.

V = 71.7 g Fe x (1 cm^{3}/ 7.87 g Fe) = 9.11 cm^{3}

**Problem #7:** Suppose you are given two cylindrical rods, one aluminum and the other tin, with identical external dimensions. What fraction of the tin rod would have to be hollow in order to give the same average density for both rods? (The density of tin is 7.31 g/cm^{3} and for aluminum, it is 2.70 g/cm^{3}.)

**Solution #1:**

Let us assume 1.00 cm^{3}of Al is present.That means 2.70 g of Al is present.

In the rod of tin that is hollow, the mass of tin present weighs 2.70 g. We know this because the average densities of the two rods are equal. Remember, the average density of the tin rod includes the empty space.

What volume of solid tin is required to provide 2.70 g? The answer is 0.370 cm

^{3}(from 2.70 g divided by 7.31 g/cm^{3}).The rest of the tin rod must be hollow. This is 0.670 cm

^{3}. Since the overall volume was 1.00 cm^{3}, the fraction of the tin rod that is hollow is 0.670

**Solution #2:**

Identical external dimensions means overall volume is the same for both rods. Let's call that V.It must be that mass Al = mass Sn.

So, let's convert from volume Al to volume Sn with equivalent mass:

2.70 g Al 1 g Sn 1 cm ^{3}V x ––––––– x ––––––– x ––––––– = 0.370V 1 cm ^{3}1 g Al 7.31 g Sn Finally the hollow fraction = volume hollow / volume total

(V − 0.370V) / V1 − 0.370 = 0.630

**Problem #8:** Osmium is one of the densest elements known. The standard density is 22.59 kg/L. What is the density in units of g/cm^{3}?

**Solution:**

Kilograms will be changed to grams by multiplying the numerator by 1000. Liters will be changed to mL by multiplying the denominator by 1000. Those two multiplications cancel each other out. And one mL equals one cm^{3}.The answer is 22.59 g/cm

^{3}.Here's a dimensional analysis set up showing the solution:

22.59 kg 1000 g 1 L 1 mL ––––––– x –––––– x ––––––– x ––––– = 22.59 g/cm ^{3}1 L 1 kg 1000 mL 1 cm ^{3}By the way, the standard density is not expressed in units of kg/L. The standard unit for density, according to the IUPAC, is kg/m

^{3}. The ChemTeam did not edit out the word standard in the question (he didn't write the original question), but decided to keep the word so as to make this comment.The standard density for osmium is 22,590 kg/m

^{3}. It is not often used, the chemical world much preferring to use the 22.59 g/cm^{3}value.

**Problem #9:** A gold wire has a diameter of 1.000 mm. What length of this wire contains exactly 1.000 mol of gold? (The density of gold is 19.32 g cm¯^{3}.)

**Solution:**

One mole of gold weighs 196.9665 g.V = mass / density = 196.9665 g / 19.32 g cm¯

^{3}= 10.195 cm^{3}The formula for volume of a cylinder is V = πr

^{2}LThe length (L) of the wire is therefore:

L = 10.195 cm

^{3}/ [(3.14159) (0.05000 cm)^{2}] = 1298 cmBy the way, note the 0.05 cm for the radius. The wire was 1 mm in diameter, which is 0.1 cm. And then half of that for the radius is 0.05 cm.

**Problem #10:** The density of lead is 11.34 g/cm^{3}. Which of the following contains the greatest mass of lead: 0.50 kg or 0.050 liter?

**Solution comparing masses:**

0.050 L = 50. mL = 50. cm^{3}(11.34 g/cm

^{3}) (50. cm^{3}) = 567 g0.50 kg = 500 g

The 0.050 L contains more mass.

**Solution comparing volumes:**

0.50 kg = 500 g500 g / 11.34 g/cm

^{3}= 44.1 cm^{3}44.1 cm

^{3}= 44.1 mL = 0.0441 LThe larger volume has the greater mass since both samples are lead. The answer to the problem is 0.050 L.

**Bonus Problem:** What is the volume in cubic inches (in^{3}) of a 0.500 lb sample of Pb?

**Short Solution:**

1) Look up the density of lead in pounds per cubic inch to find a value of 0.4098 lb/in^{3}.

2) Divide 0.500 lb by density:

0.500 lb / 0.4098 lb/in^{3}= 1.22 in^{3}

**Long Solution, deliberately starting with metric density:**

1) Look up the density of lead to find a value of 11.34 g/cm^{3}.

2) Determine how many cubic centimeters are in one cubic inch:

1 inch = 2.54 cm(1.00 in)

^{3}= (2.54 cm)^{3}<--- a cube that is one inch in each dimension is also 2.54 cm on a side1.00 in

^{3}= 16.387 cm^{3}

3) Determine the density of lead in g/in^{3}:

11.34 g 16.387 cm ^{3}–––––– x –––––––– = 185.828 g/in ^{3}cm ^{3}in ^{3}

4) We know that 1.00 pound = 453.592 g. Therefore:

0.500 lb = 226.796 g

5) What is the volume (in cubic inches) of 226.796 g of lead?

1 in ^{3}226.796 g x –––––––– = 1.22 in ^{3}185.828 g

Twenty Examples | Probs #11-25 | Probs #26-50 | All the examples & problems, no solutions | Significant Figures Menu |