Problems 26-50

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**Problem #26:** A graduated cylinder is filled to the 40.00 mL mark with mineral oil. The masses of the cylinder before and after the addition of mineral oil are 124.966 g and 159.446 g. In a separate experiment, a metal ball bearing of mass 18.713 g is placed in the cylinder and the cylinder is again filled to the 40.00 mL mark with the mineral oil. The combined mass of the ball bearing and mineral oil is 50.952 g. Calculate the density of the ball bearing.

**Solution:**

1) Determine the density of the mineral oil:

159.446 g − 124.966 g = 34.480 g34.480 g / 40.00 mL = 0.8620 g/mL

2) Determine the volume of the ball bearing:

50.952 g − 18.713 = 32.239 g (this is the mass of mineral oil)32.239 g / 0.8620 g/mL = 37.40 mL (this is the volume of mineral oil)

40.00 mL − 37.40 mL = 2.60 mL

3) The density of the ball bearing is 7.197 g/mL. This came from 18.713 g divided by 2.60 mL.

**Problem #27:** A calibrated flask was filled to the 25.00 mark with ethyl alcohol. By weighing the flask before and after adding the alcohol, it was determined that the flask contained 19.7325 g of alcohol. In a second experiment, 25.9880 g of metal beads were added to the flask, and the flask was again filled to the 25.00 ml mark with ethyl alcohol. The total mass of the metal plus alcohol in the flask was determined to be 38.5644 g. What is the density of the metal in g/mL?

**Solution:**

1) Determine density of alcohol:

19.7325 g / 25.00 mL = 0.7893 g/mL

2) Determine volume of metal beads:

38.5644 g − 25.9880 g = 12.5764 g (this is the mass of alcohol)12.5764 g / 0.7893 g/mL = 15.9336 mL (this is the volume of alcohol)

25.00 mL − 15.9336 mL = 9.0664 mL (this is the volume of the metal beads)

3) Density of the metal:

25.9880 g / 9.0664 mL = 2.8664 g/mL

**Problem #28:** A bar of magnesium metal attached to a balance by a fine thread weighed 31.13 g in air and 19.35 g when completely immersed in hexane (density = 0.659 g/cm^{3}). Calculate the density of this sample of magnesium.

**Solution:**

1) Calculate the apparent loss of mass due to immersion:

31.13 g − 19.35 g = 11.78 g

2) Calculate the volume of hexane displaced:

11.78 g / 0.659 g/cm^{3}= 17.876 cm^{3}

3) Calculate the density of magnesium:

31.13 g / 17.876 cm^{3}= 1.74 g/cm^{3}(to three sig figs)

**Problem #29:** Brass is a zinc and copper alloy. What is the mass of brass in a brass cylinder that is 1.2 in long and a diameter of 1.5 in if the brass is made of 67% copper by mass and 33% zinc by mass? The density of copper is 8.94 g/cm^{3} and 7.14 g/cm^{3} of zinc? Assume brass varies linearly with composition.

**Solution:**

a) Convert inches to centimeters

1.2 in x 2.54 cm/in = 3.048 cm

1.5 in x 2.54 cm/in = 3.81 cm

b) Determine volume of cylinder:

V = πr^{2}h = (3.14159) (1.905 cm)^{2}(3.048 cm)

V = 34.75 cm^{3}

c) Determine mass of copper and mass of zinc in cylinder as if each were 100%:

copper:x / 34.75 cm

^{3}= 8.94 g/cm^{3}

x = 310.665 gzinc:

y/ 34.75 cm

^{3}= 7.14 g/cm^{3}

y = 248.115 g

d) However, Cu is 67% of the mass and Zn is 33%:

Cu: (310.665 g) (0.67) = 208.14555

Zn: (248.115 g) (0.33) = 81.878 g

e) The mass of the brass cylinder is:

208.14555 + 81.878 = 290.02 g; to two sig figs (which seems reasonable) the answer is 290 g

**Problem #30:** The density of osmium is 22.57 g/cm^{3}. If a 1.00 kg rectangular block of osmium has two dimensions of 4.00 cm x 4.00 cm, calculate the third dimension of the block.

## Find the Missing Dimension of a Block of Osmium

**Problem #31:** A 15.8 g object was placed into an open container that was full of ethanol. The object caused some ethanol to spill, then it was found that the container and its contents weighed 10.5 grams more than the container full of ethanol only. What is the density of the object?

**Solution:**

a) Let x = the mass of ethanol in the full, open container. Therefore:

x + 15.8 = the mass of the full container plus the object before any spilling.

x + 10.5 = the mass of the full container plus the object after it was dropped in.

Please note that the x + 10.5 is not the mass after some ethanol has spilled out. Notice that the problem says "10.5 grams more than the container __ full__ of ethanol only."

b) The difference is the mass of ethanol that spilled out:

(x + 15.8) − (x + 10.5) = 5.3 g

c) Determine the volume of 5.3 g of ethanol:

5.3 g / 0.789 g/mL = 6.72 mL0.789 g/mL is the density of ethanol.

d) Determine the density of the object:

15.8 g / 6.72 mL = 2.35 g/mLSince we will assume the object is solid, let us write the answer as 2.35 g/cm

^{3}

**Problem #32:** A sheet of aluminum foil measures 30.5 cm by 75.0 cm and has a mass of 9.94 g. What is the thickness of the foil? (The video solves a similar problem to this one. It is not a solution to problem #32.)

## Determine the thickness of aluminum foil

**Problem #33:** The mass of a gold nugget is 84.0 oz. What is its volume in cubic inches? (Density of gold: 19.31 g/cm^{3}; 435.6 g = 1.00 pound; 16.0 oz = 1.00 pound; 16.387 cm^{3} = 1.00 in^{3})

**Solution:**

a) Convert 84.0 oz to grams:

84.0 oz times (1.00 pound/16.0 oz) = 5.25 pound5.25 lb times (453.6 g/1.00 pound) = 2381.4 g

b) Determine volume of 2381.4 g of gold in cm^{3}:

volume = 2381.4 g ÷ 19.31 g/cm^{3}= 123.3247 cm^{3}(kept some guard digits)

c) Convert cm^{3} to in^{3}:

123.3247 cm^{3}times (1.00 in^{3}/16.387 cm^{3}) = 7.52 in^{3}This is the answer to the problem.

The 16.387 cm 1) The cube to the right is 1.00 inch in each dimension, |

**Problem #34:** Gold can be hammered into extremely thin sheets called gold leaf. If a 204 mg piece of gold (density = 19.32 g/cm^{3}) is hammered into a sheet measuring 2.4 feet by 1.0 feet, what is the average thickness of the sheet in meters?

**Solution:**

1) Determine the volume of gold present:

0.204 g / 19.32 g/cm^{3}= 0.010559 cm^{3}

2) Convert cm^{3} to m^{3}:

We know that 1 cm^{3}= 10¯^{6}m^{3}, so:0.010559 cm

^{3}= 0.010559 x 10¯^{6}m^{3}= 1.0559 x 10¯^{8}m^{3}

3) Convert 1.0 foot and 2.4 feet to meters:

1 foot = 0.3048 m; 2.4 feet = 0.73152 m

4) Solve for the missing dimension:

(0.3048 m) (0.73152 m) (x) = 1.0559 x 10¯^{8}m^{3}x = 4.73 x 10¯

^{8}m

I used Google to make my conversions (type "convert 1 cubic centimeter to cubic meters" in the search box in Google and see what happens.)

Bonus question: How might the thickness be expressed without exponential notation, using an appropriate metric prefix?

This question does not seek this for an answer:4.73 x 10¯

^{8}m = 0.0000000473 mWhat is means is to convert the meter value to a different

namedmetric prefix, so that the numeric part is somewhere between 1 and 1000:We see that the prefix nano- means 10¯

^{9}m, so4.73 x 10¯

^{8}m = 47.3 nm

**Problem #35:** How long is a 22.0 gram piece of copper wire with a diameter of 0.250 millimeters? Density = 8.96 g/cm^{3}

**Solution:**

1) Determine volume of wire:

22.0 g / 8.96 g/cm^{3}= 2.45536 cm^{3}

2) Use the formula for volume of a cylinder:

V = πr^{2}h2.45536 cm

^{3}= (3.14159) (0.0125 cm)^{2}hh = 5.00 x 10

^{3}cmNote the radius in cm is used, not the diameter in mm.

**Problem #36:** How long is a copper wire with a diameter of 0.250 millimeters? The density of copper is 8.96 g/cm^{3} and the mass of the wire is 22.0 g.

**Solution:**

1) We know the following two equations:

D = m /V ---> where D is density, m is mass, V is volumeV = πr

^{2}L ---> V is volume, r is the radius, and L is the length of the wire.

2) Put them together:

m D = ––– πr ^{2}L

3) Solve for L:

m L = ––– πr ^{2}D

4) Since density uses cm, the radius must be in cm:

0.250 mm = 0.0250 cm, and the radius is 0.0125 cm

5) Solve for L:

22.0 g L = –––––––––––––––––––––––––––––– (3.14159) (0.0125 cm) ^{2}(8.96 g/cm^{3})L = 5002.0 cm

To three sig figs (and converted to meters), 50.0 m

**Problem #37:** A 12.0 cm long cylindrical glass tube, sealed at one end is filled with ethanol. The mass of ethanol needed to fill the tube is found to be 9.60 g. The density of ethanol is 0.789 g/mL. What is the inner diameter of the tube in centimeters?

**Solution:**

1) Determine the volume of ethanol:

9.60 g / 0.789 g/mL = 12.1673 mL

2) Use the formula for volume of a cylinder:

V = πr^{2}h12.1673 cm

^{3}= (3.14159) (r^{2}) (12.0 cm)r

^{2}= 0.322748r = 0.568 cm

Note the change from mL to cm

^{3}

**Problem #38:** A 23.200 g sample of copper is hammered to make a uniform sheet of copper with a thickness of 0.100 mm. What is the area of this sheet in cm^{2} given the density of copper to be 8.96 g/cm^{3}?

**Solution:**

1) Determine volume of Cu foil:

23.200 g / 8.96 g/cm^{3}= 2.5893 cm^{3}

2) Determine area:

(area) times 0.0100 cm = 2.5893 cm^{3}area = 259 cm

^{2}(to 3 sig figs)

**Problem #39:** A piece of copper foil has a mass of 4.924 g, a length of 3.62 cm, and a width of 3.02 cm Calculate the thickness in mm, assuming the foil has uniform thickness.

**Solution:**

1) Let us look up the density of copper on the Intertubes:

8.96 g/cm^{3}

2) Calculate volume of copper foil:

4.924 g divided by 8.96 g/cm^{3}= 0.550 cm^{3}

3) Calculate the missing dimension:

(3.62 cm) (3.02 cm) (x) = 0.550 cm^{3}x = 0.0503 cm

4) Convert cm to mm:

0.503 mm

**Problem #40:** A 50.00 g block of wood shows an apparent mass of 5.60 g when suspended in water at 20.0 °C water (the density of water at 20.0 °C is 0.99821 g/mL). What is the density of the block?

**Solution:**

1) calculate the apparent loss of mass:

50.00 g − 5.60 g = 44.40 g

2) Using Archimedes Principle:

"the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object." SourceWe determine that 44.40 g of water have been displaced.

3) Determine the volume of 44.40 g of water at 20.0 °C:

44.40 g divided by 0.99821 g/mL = 44.48 mL

4) Determine the density of the block:

50.00 g / 44.48 mL = 1.124 g/mLSince densities of solid are generally reported in g/cm

^{3}(amd 1 mL = 1 cm^{3}), we report the final answer as:1.124 g/cm^{3}

**Problem #41:** What is the mass of a flask filled with acetone (d = 0.792 g/cm^{3}) if the same flask filled with water (d = 1.000 g/cm^{3}) weighs 75.20 gram? The empty flask weighs 49.74 g.

**Solution:**

1) Determine mass of water:

75.20 g − 49.74 g = 25.46 g

2) Determine volume of water:

25.46 g divided by 1.000 g/cm^{3}= 25.46 cm^{3}

3) Determine mass of acetone:

25.46 cm^{3}times 0.792 g/cm^{3}= 20.16432 g

4) Determine mass of flask + acetone:

20.16432 g + 49.74 g = 69.90 g (to four sig figs)

**Problem #42:** A container is filled with water at 20.0 °C, just to an overflow spout. A cube of wood with edges of 1.00 in. is submerged so its upper face is just at the level of water in the container. When this is done, 10.8 mL of water is collected through the overflow spout. Calculate the density of the wood.

**Solution:**

1) By Archimedes Principle:

the wood block weighs 10.8 g

2) One cubic inch equals:

16.387 cm^{3}

3) Calculate the density of the wood:

10.8 g / 16.387 cm^{3}= 0.659 g/cm^{3}

**Problem #43:** A pycnometer is a device used to determine density. It weighs 20.578 g empty and 31.609 g when filled with water (density = 1.000 g/cm^{3}). Some pieces of a metal are placed in the empty, dry pycnometer and the total mass is 44.184 g. Water is then added to exactly fill the pycnometer and the total mass is determined to be 54.115 g. What is the density of the metal?

**Solution:**

1) Determine mass of metal:

44.184 − 20.578 = 23.606 g

2) Determine mass, then volume of water to fill up pycnometer:

54.115 g − 44.184 g = 9.931 gThis mass of water is equal to 9.931 cm

^{3}

3) Determine volume of pycnometer:

31.609 g − 20.578 g = 11.031 gThe volume of the pycnometer is 11.031 cm

^{3}

4) Determine volume of metal:

11.031 cm^{3}− 9.931 cm^{3}= 1.100 cm^{3}

5) Determine density of metal:

23.606 g / 1.100 cm3 = 21.460 g/cm^{3}Although the problem does not ask for the identity of the metal, the value determined is the density of platinum.

**Problem #44:** Vinaigrette salad dressing consists mainly of oil and vinegar. The density of olive oil is 0.918 g/mL, the density of vinegar is 1.006 g/mL, and the two do not mix. If a certain mixture of olive oil and vinegar has a total mass of 402.3 g and a total volume of 421.0 mL, what is the volume of oil and what is the volume of vinegar in the mixture?

**Solution:**

a1) first equation concerns the olive oil:

x/y = 0.918 g/mL

a2) second equation concerns the vinegar:

(402.3 − x) / (421.0 − y) = 1.006 g/mL

b) rearrange the olive oil equation:

x = 0.918y

c) substitute into the vinegar equation and solve for y:

(402.3 − 0.918y) / (421.0 − y) = 1.006423.526 − 1.006y = 402.3 − 0.918y

21.226 = 0.088y

y = 241.2 mL (this is the volume of the olive oil)

d) solve for the volume of vinegar:

421.0 − 241.2 = 179.8 mL

This type of problem (with slightly different numbers), done as a video.

**Problem #45:** What is the radius of 0.663 g of copper wire (density = 8.96 g/cm^{3}) that is 22.6 cm in length?

**Solution:**

1) Determine the volume of the wire:

Density = mass / volumerearranging:

Volume = mass / density

V = 0.663 g / 8.96 g/cm

^{3}= 0.0739955 cm^{3}

2) Use the formula for volume of a cylinder to determine the radius:

V = πr^{2}hRearranging:

V r ^{2}=––– πh

0.0739955 cm ^{3}r ^{2}=–––––––––––––––––– (3.14159) (22.6 cm) r

^{2}= 0.00104219 cm^{2}r = 0.0323 cm (to three sig figs)

**Problem #46:** A thin coating of gold was placed onto a tray that measured 277.5 mm by 142.5 mm. The mass of the tray increased by 0.0624 g during the process. The density of gold is 19.32 g/cm^{3}. (a) Calculate the thickness of the plating. (b) Determine the approximate number of gold atoms in the thin coating. (c) How many atoms of gold are in the thickness of the layer?

**Solution:**

1) Let us suppose the tray is plated on only one side. Determine the volume of gold that was deposited on the plate:

0.0624 g / 19.32 g/cm^{3}= 0.0032298 cm^{3}

2) Convert the given mm values to cm:

(277.5 mm) (1 cm / 10 mm) = 27.75 cm

(142.5 mm) (1 cm / 10 mm) = 14.25 cm

3) Determine the thickness of the gold layer (answer to (a)):

0.0032298 cm^{3}= (27.75 cm) (14.25 cm) (x)x = 8.168 x 10¯

^{6}cm

4) Determine number of atoms in the thin layer (answer to (b)):

0.0624 g / 196.96657 g/mol = 0.000316805 mol(0.000316805 mol) (6.022 x 10

^{23}atoms/mol) = 1.908 x 10^{20}atoms (to four sig figs)

5) To determine how many atoms thick, we will make a simplifying assumption that the atoms stack one on top of the other.

6) We need to know the diameter of a gold atom. We look up the empirical atomic radius of gold to find 135 pm, so the diameter is 270 pm. Convert this value to cm:

(270 pm) (1 cm / 10^{10}pm) = 2.70 x 10¯^{8}cm

7) Divide the thickness of the gold layer by the diameter of a gold atom:

0.0000081676624 cm / 2.70 x 10¯^{8}cm/atom = 302 atoms (answer to (c))

8) A more realistic modeling of gold atoms in a layer would use gold's crystalline structure of body-centered cubic. This answer discusses that.

**Problem #47:** Mercury is often used as an expansion medium in a thermometer. The mercury sits in a bulb on the bottom of the thermometer and rises up a thin capillary as the temperature rises. Suppose a mercury thermometer contains 3.400 g of mercury and has a capillary that is 0.1900 mm in diameter

How far does the mercury rise in the capillary when the temperature changes from 0.0 °C to 25.0 °C? The density of mercury at these temperatures is 13.596 g/cm^{3} and 13.534 g/cm^{3}, respectively.

**Solution:**

1) Volume of Hg at 0.0 °C:

3.400 g / 13.596 g/cm^{3}= 0.250074 cm^{3}

2) Volume of Hg at 25.0 °C:

3.400 g / 13.534 g/cm^{3}= 0.251219 cm^{3}

3) Volume difference:

0.251219 − 0.250074 = 0.001145 cm^{3}

4) Change 0.1900 mm to cm:

(0.190 mm) (1 cm / 10 mm) = 0.01900 cm

5) Use the formula for volume of a cylinder:

V = (4/3) π r^{2}h0.001145 cm

^{3}= (4/3) (3.14159) (0.009500 cm)^{2}(h)h = 3.029 cm (to 4 sig figs)

**Problem #48:** A container (open to the atmosphere) is filled to a volume of 1.40 L with water at 25 °C and then frozen to −10 °C. What new volume does the ice occupy? Water has a density of 0.997 g/cm^{3} at 25 °C; ice has a density of 0.917 g/cm^{3} at −10 °C

**Solution:**

1) I'm going to give you the entire set up first, then comment on it below:

1000 mL 1 cm ^{3}0.997 g 1 cm ^{3}1 mL 1 L 1.40 L x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– x ––––––– = 1.52 L (to three sig figs) 1 L 1 mL 1 cm ^{3}0.917 g 1 cm ^{3}1000 mL

2) Look at the section that ends at the 0.997. Here's what's happening:

first conversion ---> L to mL

second conversion ---> mL to cm^{3}

third conversion ---> cm^{3}to gramsIf you do those calculations, you get 1395.8 g. That's the mass of water in the container. The mass will not change as the water turns to ice.

3) The very next conversion (the one involving 0.917) is a division. The mass is divided by the density of the ice to give the volume of the ice in cm^{3}

4) The last two conversions are the reverse of the first two:

next-to-last ---> convert cm^{3}to mL

last ---> convert mL to L

5) Here is a different solution:

0.997 g 1 cm ^{3}1.40 L x ––––––– x ––––––– = 1.52 L 1 cm ^{3}0.917 g And it, of course, works perfectly, with all the units cancelling properly. It's a streamlined version of the long one I gave at the beginning of the answer. Is it better or worse than the long one up above? The answer, of course, is neither. You, as the student, would be well-served to be familiar with what is happening in both approaches.

**Problem #49:** Rolls of aluminum foil are 306 mm wide and 0.0140 mm thick. What maximum length of aluminum foil can be made from 0.834 kg of aluminum? (The density of Al is 2.70 g/cm^{3})

**Solution:**

1) The maximum length is, of course, the third side of the volume that the foil occupies. Let's get the volume of the foil:

(0.834 kg) (1000 g / kg) = 834 g834 g / 2.70 g/cm

^{3}= 308.889 cm^{3}

2) The width and thickness are given in mm. Convert each to cm:

(306 mm) (1 cm / 10 mm) = 30.6 cm

(0.0140 mm) (1 cm / 10 mm) = 0.00140 cm

3) Determine the length of foil:

(30.6 cm) (0.00140 cm) (x) = 308.889 cm^{3}x = 7210.294 cm

4) Although no specified unit is given for the answer, let's convert it to meters (and round off to three sig figs):

(7210.294 cm) (1 m / 100 cm) = 72.1 m

**Problem #50:** What would be the volume in cubic inches of a sample of aluminum with a mass of 250.5 grams?

**Solution #1:**

1) Determine volume of sample:

250.5 g / 2.70 g/cm^{3}= 92.77778 cm^{3}

2) Think of 92.77778 cm^{3} like this:

92.77778 cm x 1 cm x 1 cm

3) There is an exact conversion between cm and inch. It is this:

2.54 cm = 1 inch.4) Convert cm

(92.77778 cm x 1 in/2.54 cm) x (1 cm x 1 in/2.54 cm) x (1 cm x 1 in/2.54 cm) = 5.66 in^{3}

**Solution #2:**

1) Determine volume of sample:

250.5 g / 2.70 g/cm^{3}= 92.77778 cm^{3}

2) Convert 1 cubic inch to cubic centimeters:

1 in = 2.54 cm1 in

^{3}= (1 in) (1 in) (1 in)1 in

^{3}= (2.54 cm) (2.54 cm) (2.54 cm)1 in

^{3}= 16.387064 cm^{3}

3) Convert volume of Al in cm^{3} to its equivalent volume in in^{3}:

1 in ^{3}92.77778 cm ^{3}x ––––––– = 5.66 in ^{3}16.387064 cm ^{3}

**Bonus Problem:** The density of a liquid is determined by successively weighing a graduated cylinder containing 10, 20, 30, 40 and 50 mL. If the volume is plotted along the vertical axis and the total weight of the cylinder and the liquid is plotted on the horizontal axis:

(a) the slope of the line is the inverse of the density of the liquid

(b) none of the above statements are correct

(c) the slope of the line is 1.0

(d) the line will pass through the (0,0) origin

(e) the intercept on the x axis is the value of the weight of the cylinder

Comment: I copied the answer below from an "answers" website. The writer makes a mistake and identifies, incorrectly, answer (e) as the correct answer. He then corrects his mistake. I have decided to keep the mistaken answer.

**Solution:**

The correct answer is (e).Why? At zero volume of the liquid, the mass will be that of the empty cylinder.

Answer (a) is a cute one because volume on the y-axis and mass is on the x-axis. That means the slope of the line (change in y over change in x) wold be the inverse of density (which is mass over volume). However, keep in mind that the the values of mass include a constant value, the mass of the empty cylinder. If we were to plot just the mass of the liquid, we would make answer a correct. However, including the mass of the cylinder (a constant value) renders answer a not true.

Answer (a) is true, I made a mistake. The slope of the line is the same, even if it is shifted. When you calculate the slope using the distance formula, the constant value of the weight of the cylinder will subtract out.

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