Question #4e810

1 Answer
Apr 16, 2016

Answer:

#"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.