Question 2.17: UNIT CONVERSION A chemist needs a 23.5-kg sample of ethanol ......

• Scan the problem for one or more numbers and their associated units.
  This number (or numbers) is (are) the starting point(s) of the calculation. Write them down as given.
• Scan the problem to determine what you are asked to find. Sometimes the units of this quantity are implied; other times they are specified. Write down the quantity and/or units you are asked to find.

density = 0.789 g/cm^3

FIND: volume in L

• For problems involving only conversions, focus on units. The solution map shows how to get from the units in the given quantity to the units in the quantity you are asked to find.
• For problems involving equations, focus on the equation. The solution map shows how the equation takes you from the given quantity (or quantities) to the quantity you are asked to find.
• Some problems may involve both unit conversions and equations, in which case the solution map employs both of the above points.

RELATIONSHIPS USED

0.789 g/cm^3 (given in problem)

1000 g = 1 kg (Table 2.2)

1000 mL = 1 L (Table 2.2)

1 mL = 1 cm^3 (Table 2.3)

•  For problems involving only conversions, begin with the given quantity and its units. Multiply by the appropriate conversion factor(s), canceling units, to arrive at the quantity you are asked to find.
• For problems involving equations, solve the equation to arrive at the quantity you are asked to find. (Use algebra to rearrange the equation so that the quantity you are asked to find is isolated on one side.) Gather each of the quantities that must go into the equation in the correct units. (Convert to the correct units using additional solution maps if necessary.) Finally, substitute the numerical values and their units into the equation and calculate the answer.
• Round the answer to the correct number of significant figures. Use the significant figure rules from Sections 2.3 and 2.4.

23.5  \cancel{ kg} \times \frac{1000  \cancel{g}}{1  \cancel{ kg}} \times \frac{1  \cancel{cm^3}}{0.789  \cancel{g}} \times \frac{1  \cancel{mL}}{1  \cancel{cm^3}} \times \frac{1  L}{1000  \cancel{mL}} = 29.7845 L

29.7845 L = 29.8 L

• Does the magnitude of the answer make physical sense? Are the units correct?

= 1.094 yards (yd)

= 1000 cubic centimeters (cm^3)