Problems #1-10

**Problem #1:** Convert 0.500 cubic feet to liters

**Solution:**

1) Let's set up the dimensional analysis first and then comment on it:

(12 inch) ^{3}(2.54 cm) ^{3}1 mL 1 L 0.5 ft ^{3}x ––––––– x ––––––– x ––––– x ––––––– = 14.2 L 1 ft ^{3}(1 in) ^{3}1 cm ^{3}1000 mL

2) The conversions:

first ---> convert cubic feet to cubic inches

second ---> convert in^{3}to cubic centimeters

third ---> convert cm^{3}to mL

fourth ---> convert mL to L

3) Comment:

Note the (12 inch)^{3}. You must remember to "distribute" the cube. While it is true that 12 inches equals 1 foot, you have to remember that 12 in^{3}DOES NOT equal 1 ft^{3}. You have to cube the 12, as in:1 ft^{3}= (12 in) (12 in) (12 in) = 1728 in^{3}The number 12 AND the unit inch both get cubed.

**Problem #2:** Convert 0.500 cubic feet to dm^{3}

**Solution:**

1) The dimensional analysis set up:

(12 inch) ^{3}(2.54 cm) ^{3}(1 dm) ^{3}0.5 ft ^{3}x ––––––– x ––––––– x ––––––– = 14.2 dm ^{3}1 ft ^{3}(1 in) ^{3}(10 cm) ^{3}

2) The conversions:

first ---> convert cubic feet to cubic inches

second ---> converts in^{3}to cubic centimeters

third ---> converts cm^{3}to dm^{3}

3) Comment:

There are 10 cm in 1 dm. Remember, deci- means 0.1 while centi- means 0.01. There are 1000 cm^{3}in 1 dm^{3}, just like there are 1000 mL in 1 L.

**Problem #3:** Convert the density of gold, 19.3 g/cm^{3}, to kg/L.

**Solution:**

1) Convert grams to kg:

19.3 g/cm^{3}times (1 kg / 1000 g) = 0.0193 kg/cm^{3}

2) Convert cm^{3} to mL:

0.0193 kg/cm^{3}times (1 mL / 1 cm^{3}) = 0.0193 kg/mL

3) Convert mL to L:

0.0193 kg/mL times (1000 mL / L) = 19.3 kg/L

4) Often, a teacher will want you to write out the conversions in one long line:

19.3 g/cm^{3}x (1 kg / 1000 g) x (1 mL / 1 cm^{3}) x (1000 mL / L) = 19.3 kg/L

**Problem #4:** Change 1.0 kg/cm^{3} to g/mm^{3}

**Solution:**

(1.0 kg / cm^{3}) x (1000 g / kg) = 1000 g/cm^{3}(1000 g / cm

^{3}) x (1 cm^{3}/ 1000 mm^{3}) = 1.0 g / mm^{3}Remember, 10 mm in one cm, so 10 mm x 10 mm x 10 mm equals 1000 mm

^{3}in one cm^{3}.Teachers like these conversions where you start with 1 and end with 1. Students don't expect it and think they have done something wrong. The video has another example.

## Convert kg/m

^{3}to g/L

**Problem #5:** Convert 25.0 mL to mm^{3} using dimensional analysis.

**Solution via cm ^{3}:**

1 cm ^{3}10 ^{3}mm^{3}25.0 mL x ––––– x ––––––– = 2.50 x 10 ^{4}mm^{3}1 mL 1 cm ^{3}The 1 mL equals 1 cm

^{3}conversion is very handy.There are 10 mm in every cm, so 10 cubed is 10

^{3}.

**Solution via dm ^{3}:**

1 L 1 dm ^{3}10 ^{6}mm^{3}25.0 mL x ––––– x ––––– x ––––––– = 2.50 x 10 ^{4}mm^{3}10 ^{3}mL1 L 1 dm ^{3}The 1 L equals 1 dm

^{3}conversion is very handy.There are 100 mm in every dm, so 100 cubed is 10

^{6}

**Problem #6:** Convert 5.51 g/cm^{3} to lb/ft^{3}.

**Solution:**

1) Do the non-cubic conversion first:

1 lb 5.51 g/cm ^{3}x–––––– = the answer 453.6 g I'm not going to work it out yet, I simply wanted to show the above conversion. I'll leave the working out until the cm

^{3}to ft^{3}is added in.

2) We can look up a direct conversion from cm^{3} to ft^{3} (you may, I will not bother). The usual technique (which is what I follow) is to go via common conversions, ones that teachers tend to have their students memorize. In this specific case, we would go from cm^{3} to cubic inch (since 1 inch exactly equals 2.54 cm) and then from in^{3} to cubic foot (because 1 foot exactly equals 12 inches).

1 lb (2.54 cm) ^{3}(12 in) ^{3}5.51 g/cm ^{3}x–––––– x –––––––– x –––––––– = 344 lb/ft ^{3}453.6 g (1 in) ^{3}(1 ft) ^{3}Notice the style I used for the cubic units: (1 ft)

^{3}, (2.54 cm)^{3}, and so on. Sometimes, you see it done without the parentheses: 1 ft^{3}or 2.54^{3}cm^{3}might be what you woud see. Notice how the cube on the numeral 1 is eliminated.5.51 g/cm

^{3}is the density of the Earth.

**Problem #7:** How many 1 cm cubes would it take to construct a cube that is 4 cm on edge?

**Solution:**

1) The formula for volume of a cube is:

V = l*w*h

2) Insert 4 cm for each of the three dimensions:

V = 4 cm times 4 cm times 4 cm = 64 cm^{3}Sixty-four 1 cm cubes would be required.

**Problem #8:** Convert 6.230 x 10¯^{2} kg/mm^{3} to g/L

**Solution:**

1) You have to do a kg to g conversion and then take mm^{3} to cm^{3} to mL to L. Here it is:

0.06230 kg 1000 g (10 mm) ^{3}(1 cm) ^{3}1000 mL ––––––––– x ––––– x ––––––– x ––––––– x ––––––– = 6.230 x 10 ^{7}g/Lmm ^{3}1 kg (1 cm) ^{3}1 mL 1 L

2) Notice that kg/mm

**Problem #9:** If a box measures 3.00 cm wide, 50.5 mm long and 0.520 m high, what is its volume in cubic feet?

**Solution:**

1) One of the common pathways from metric units to English units is 2.54 cm = 1 inch. Let's take each measurement to cm and then calculate the volume in cm^{3}:

(50.5 mm) (1 cm / 10 mm) = 5.05 cm

(0.520 m) (100 cm / 1 m) = 520. cmvolume ---> (3.00 cm) (5.05 cm) (520. cm) = 7878 cm

^{3}

2) Convert from cm^{3} to cubic inches:

(7878 cm^{3}) (1 inch / 2.54 cm)^{3}= 480.745 in^{3}Notice how I cubed the entire conversion factor rather than cubing them individually, as in (1 in)

^{3}/ (2.54 cm)^{3}

3) Convert from cubic inches to cubic feet:

(480.745 in^{3}) (1 foot / 12 inches)^{3}= 0.278 ft^{3}(to three sig figs)

**Problem #10:** Convert 1 m^{3} to dm^{3} and then to L

**Solution:**

(1 m)^{3}(10 dm / 1 m)^{3}= 1000 dm^{3}Note how I wrote 1 m

^{3}Since 1 dm

^{3}= 1 L, 1000 dm^{3}= 1000 LAn alternate way to write the above conversion:

(1 m

^{3}) (10^{3}dm^{3}/ 1^{3}m^{3}) = 1000 dm^{3}

**Bonus Problem #1:** Convert 2.70 g/cm^{3} to kg/m^{3} then to lb/ft^{3}

**Solution:**

1) Convert grams to kilograms:

1 kg 2.70 g / cm ^{3}x ––––– = 0.00270 kg / cm ^{3}1000 g

2) Convert kg/cm^{3} to kg/cm^{3}:

(100 cm) ^{3}0.00270 kg / cm ^{3}x –––––––– = 2.70 x 10 ^{3}kg / m^{3}(1 m) ^{3}

3) Done in dimensional analysis style:

2.70 g 1 kg (100 cm) ^{3}––––––– x ––––––– x ––––––– = 2.70 x 10 ^{3}kg / m^{3}1 cm ^{3}1000 g (1 m) ^{3}

4) To continue on, we need to look up the kg-to-pound conversion (1 kg = 2.2046226 lb) and the meter-to-foot conversion (1 m = 3.28084 ft). Here is the full conversion in dimensional analysis style:

2.70 g 1 kg (100 cm) ^{3}(1 m) ^{3}2.2046226 lb ––––––– x ––––––– x ––––––– x –––––––––– x ––––––––––– = 168 lb / ft ^{3}(to three sig figs)1 cm ^{3}1000 g (1 m) ^{3}(3.28084 ft) ^{3}1 kg

**Bonus Problem #2:** The German chemist Fritz Haber proposed paying off the reparations imposed against Germany after World War I by extracting gold from seawater. Given the following data, what was the dollar amount of the gold that could have been extracted from 3.00 m^{3} of seawater? (The price of gold at the time was $0.68 per gram of gold. Gold occurs in seawater to the extent of 5.15 x 10^{11} atoms per gram of seawater. The density of seawater is 1.0273 g/cm^{3}.)

**Solution:**

1) Convert 3.00 m^{3} to cm^{3}:

(100 cm) ^{3}3.00 m ^{3}x ––––––– = 3.00 x 10 ^{6}cm^{3}1 m ^{3}

2) Determine the mass of sea water in 3.00 x 10^{6} cm^{3}:

(3.00 x 10^{6}cm^{3}) (1.0273 g/cm^{3}) = 3.0819 x 10^{6}g of sea water

3) Determine atoms of gold present:

(3.0819 x 10^{6}g) (5.15 x 10^{11}atoms/g) = 1.58718 x 10^{18}atoms of gold

4) Determine moles of gold:

1.58718 x 10^{18}atoms of gold / 6.022 x 10^{23}atoms/mole = 2.635636 x 10¯^{6}mole

5) Determine grams of gold:

(2.635636 x 10¯^{6}mole) (196.96657 g/mol) = 0.000519132 g of gold

6) Determine the dollar amount:

(0.68 dollar/g) (0.000519132 g) = $0.000353

7) Dimensional analysis, baby!!

100 ^{3}cm^{3}1.0273 g 5.15 x 10 ^{11}atoms1 mol 196.96657 g $0.68 3.00 m ^{3}x––––––– x ––––––– x ––––––––––––––– x ––––––––––––––– x –––––––––– x ––––– = $0.000353 1 m ^{3}1 cm ^{3}1 g 6.022 x 10 ^{23}atoms1 mol 1 g

8) Let's reverse the problem: how many cubic meters of sea water would be required to supply 1.00 dollar of gold? For this problem, I will provide only the dimensional analysis set up. Notice that it is the reverse of the previous problem.

1 g 1 mol 6.022 x 10 ^{23}atoms1 g 1 cm ^{3}1 m ^{3}$1.00 x ––––– x –––––––––– x ––––––––––––––– x –––––––––––––– x ––––––– x –––––––– = 8498 m ^{3}$0.68 196.96657 g 1 mol 5.15 x 10 ^{11}atoms1.0273 g 100 ^{3}cm^{3}