Question #4e810

1 Answer
Apr 16, 2016

"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.