Assignment 02 ANSWERS
Math Review: Measurement and Conversions
INSTRUCTIONS (ITEMS 1-12): All of these items are typical single step conversions; the equivalence or
the conversion factor is provided for each of the items. Write the equivalence equation or conversion
factor and then solve the problem.
Convert 3.82 gal to L: gal = 3.785 L
1. Conversion factor: 3.785 L/gal
2. Solution: 3.82 gal x 3.785 L
= 14.4587 L = 14.5 L
gal
Convert 18.2 kilogram to grams: kg = 103 g
3. Conversion factor: 103g
kg
4. Solution: 18.2 kg x 103 g = 18.2 x 103 g =1.82 x 104g
kg
Convert 16.80 fluid ounces to milliliters: oz = 29.57 mL
3. Conversion factor: 29.57 ml/oz
4. Solution: 16.80 fl.oz. x 29.57 mL = 496.776 mL = 4.968 x 102 mL
fl.oz.
Convert 56 centimeters to meters:
m/102 cm
5. Equivalence equation: 102 cm = m
6. Solution: 56 cm x ___m___ = 56 x 10-2 m = 5.6 X 10-1 m or 0.56 m
102 cm
Convert 10.5 quarts to gallons: 4 qt = gal
7. Conversion factor: gal/4 qt
8. Solution: 10.5 qt x gal = 2.625 qt = 2.63 gal
qt
Convert 0.0765 L to milliliters: 103 mL = L
9. Conversion factor: 103 mL/L
10. Solution: 7.65 x 10-2 L x 103 mL = 7.65 x 10 L or 76.5 L
L
Convert 72.6 kilograms to pounds: 2.20 lb = kg
11. Conversion factor: 2.20 lb/kg
12. Solution: 72.6 kg x 2.20 lb
= 159.72 lb = 160 lb
kg
INSTRUCTIONS (ITEMS 13-16): All of these items are conversions that must be carried out by more
than one step; the equivalence equations to be used in each case are given. Solve the problem using the
equivalence equations provided to produce the necessary conversion factors. Show all calculations.
Convert 2.00 L to ounces: L = 103 mL; oz = 29.6 mL
13-14. Solution: 2.00 L x 103 mL x
L
oz
= 6.76 x 101 oz or 67.6 oz
2.96 x 101 mL
Convert 1000 cubic centimeters (cm3) to cubic inches (in3): in = 2.54 cm
13-14. Solution: (in)3 = (2.54 cm)3 = in3 = 16.387064 cm3 = 1.64 x 10 cm3
103 cm3 x
in3
.
0.6097 x 102 in3 = 6.10 x 101 in3
1.64 x 10 cm3
15-16. A 100mL graduated cylinder is filled with water several times and poured into a container. When the
container is completely filled, the weight of the water is 4.4 lb. How many times was the graduated cylinder
emptied into the container? Hint the density of water is 0.98 g/mL (ml H2O = 0.980 g). 1 lb = 454 g.
100 mL x
0.98 g x ____lb_____
mL
454 g
= 0.2159 lb
4.4 lb x
cylinder
. = 20.4 cyl
0.2159 lb
cylinder would be emptied 21 times the last time containing 40 ml
(Items 17-21) The formula for the ideal gas law is PV = nRT where P= pressure in atmospheres (atm),
V= volume in liters, n= moles, R= ideal gas constant, T= temperature in Kelvin (K).
17. Solve for n
PV = nRT;
n = PV/RT
18. Solve for P
PV = nRT;
P = nRT/V
19. Solve for V
PV = nRT;
V = nRT/P
20-21. Standard temperature and pressure (STP) is 273 K and 1 atm. The volume of one mole of a gas at STP is
22.4 L. Using the ideal gas law equation and the above value for K, P, n and V above, calculate the value of
R to three significant digits. Show all steps of your solution. Look up the value in the text to make sure
you are correct. Write the correct value here: ___________________________
PV = nRT; R = PV/nT = 1 atm * 22.4 L/n * 273 K = 0.0821 atm*L/n*K
(Items 22-25) The number of moles (n) of a substance is determined by dividing the mass (in grams) of a
substance by its formula mass (FM); thus n = g/FM.
22. Solve this expression for FM.
n = g/FM; FM*n = g; FM = g/n
23 Calculate the formula mass of a substance if 0.500 moles of a substance has a mass of 90.20 g.
FM = g/n = 90.20 g / 0.500 n = 180 g/n
24. Substitute the value for n derived in Item 17 into the equation derived in Item 22 and solve for FM.
n = PV =
g
RT
FM
;
PV(FM) = gRT;
FM = gRT =
PV
25. 1.25 L of a gas at exactly 300 K and 1 atmosphere has a mass of 5.0751 g. Using the expression for formula
mass derived in Item 24 and the value of R calculated in Item 20-21, calculate the formula mass of the gas.
FM = gRT = 5.0751 g * 0.0821 atm*L/n*K * 300 K
PV
1 atm * 1.25 L
= 99.9997704 g/n = 100 g/n