# Question #a7b85

##### 1 Answer

#### Explanation:

The first thing to do here is to make sure that we have the proper units needed to calculate the mass of gasoline required to increase the mass of the car by

#m_"gas" = "1303 kg" - "1271 kg" = "32 kg"#

Notice that the **density** of gasoline is given in *grams per cubic centimeter*, **in liters**.

One approach that we could use would be to convert the mass from *kilograms* to *grams* by using the fact that

#color(purple)(bar(ul(|color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))#

We will have

#32 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = "32000 g"#

Use the density of gasoline to find the volume in *cubic centimeters* first

#32000 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(0.73color(red)(cancel(color(black)("g")))) = "43835.6 cm"^3#

Now, to convert this to *liters*, we must use the fact that

#color(purple)(bar(ul(|color(white)(a/a)color(black)("1 L" = "1 dm"^3)color(white)(a/a)|)))" "# and#" "color(purple)(bar(ul(|color(white)(a/a)color(black)("1 dm"^3 = 10^3"cm"^3)color(white)(a/a)|)))#

In this case, you will have

#43853.6 color(red)(cancel(color(black)("cm"^3))) * (1color(red)(cancel(color(black)("dm"^3))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * "1 L"/(1color(red)(cancel(color(black)("dm"^3)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("44 L")color(white)(a/a)|)))#

The answer is rounded to two **sig figs**.