Question #4e810

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Question #4e810

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Answer 1 · 2016-04-16T20:28:03

$0.0391 L$ $"0.0391 L"$

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

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

$50.0 g \cdot {a density of 1.28 g mL}^{- 1}      \frac{1 mL}{1.28 g} = 39.1 mL$ $50.0 color(red)(cancel(color(black)("g"))) * overbrace("1 mL"/(1.28color(red)(cancel(color(black)("g")))))^(color(purple)("a density of 1.28 g mL"^(-1))) = "39.1 mL"$

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

$∣ ∣ ∣ ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ \frac{a}{a} 1 L = 10^{3} mL \frac{a}{a} ∣ ∣ - -------------- -$ $color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(a/a)|)))$

You will thus have

$39.1 mL \cdot \frac{1 L}{10^{3} mL} = ∣ ∣ ∣ ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ \frac{a}{a} 0.0391 L \frac{a}{a} ∣ ∣ - ---------- -$ $39.1 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = color(green)(|bar(ul(color(white)(a/a)"0.0391 L"color(white)(a/a)|)))$

The answer is rounded to three sig figs.

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Question #4e810

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