# Question #4e810

##### 1 Answer

#### Explanation:

A substance's **density** can be used as a *conversion factor* to go from **mass** to **volume** and vice versa. In essence, density tells you the mass of a given substance **per unit of volume**.

In your case, the density of glucose is said to be equal to *one unit of volume* of glucose, i.e.

So, if you know that you're getting **for every**

#50.0 color(red)(cancel(color(black)("g"))) * overbrace("1 mL"/(1.28color(red)(cancel(color(black)("g")))))^(color(purple)("a density of 1.28 g mL"^(-1))) = "39.1 mL"#

In order to express the volume in *liters*, use the conversion factor

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

You will thus have

#39.1 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = color(green)(|bar(ul(color(white)(a/a)"0.0391 L"color(white)(a/a)|)))#

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