Question 4e810

Apr 16, 2016

$\text{0.0391 L}$

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 ${\text{1.28 g mL}}^{- 1}$. This tells you that one unit of volume of glucose, i.e. $\text{1 mL}$, will have a mass of $\text{1.28 g}$.

So, if you know that you're getting $\text{1.28 g}$ of glucose for every $\text{1 mL}$, you can say that your $\text{50.0-g}$ 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

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

You will thus have

$39.1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = color(green)(|bar(ul(color(white)(a/a)"0.0391 L} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.