# Question a7b85

Oct 8, 2016

$\text{44 L}$

#### 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

${m}_{\text{gas" = "1303 kg" - "1271 kg" = "32 kg}}$

Notice that the density of gasoline is given in grams per cubic centimeter, ${\text{g cm}}^{- 3}$, but that the problem wants you to find the volume in liters.

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

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 kg" = 10^3"g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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.