Question

Question #4e810

Answer 1

`"0.0391 L"`

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 `"1.28 g mL"^(-1)`. This tells you that one unit of volume of glucose, i.e. `"1 mL"`, will have a mass of `"1.28 g"`.

So, if you know that you're getting `"1.28 g"` of glucose for every `"1 mL"`, you can say that your `"50.0-g"` sample will occupy

`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.