Question #76f75

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Question #76f75

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Answer 1 · 2015-02-12T09:59:36

The decanter will have a volume of $810. mL$ $"810. mL"$ .

You can actually get the volume you need by using dimensional analysis, and by using the formula for density separately. Here's how you'd go about determining the volume you need by dimensional analysis

$6.43 lbs \cdot \frac{453.59 g}{1 lbs} \cdot \frac{1 mL}{3.6 g} = 810.14 mL = 810. mL$ $"6.43 lbs" * ("453.59 g")/("1 lbs") * "1 mL"/"3.6 g" = "810.14 mL" = "810. mL"$ - rounded to three sig figs.

The first conversion factor goes from pounds to grams, while the second goes from grams to mL. The key here is to recognize that density is actually

$ρ = 3.6 g/mL = \frac{3.6 g}{1 mL}$ $rho = "3.6 g/mL" = "3.6 g"/"1 mL"$ $\to$ $->$ one mililiter weighs 3.6 g.

The other way you can solve it is by using the formula for density

$ρ = \frac{m}{V} \Rightarrow V = \frac{m}{ρ}$ $rho = m/V => V = m/(rho)$

You can use a single conversion factor to determine the weight in grams

$6.43 lbs \cdot \frac{453.59 g}{1 lbs} = 2916.5 g$ $"6.43 lbs" * ("453.59 g")/("1 lbs") = "2916.5 g"$

Then plug this value in the formula

$V = \frac{m}{ρ} = \frac{2916.5 g}{3.6 g/mL} = 810. mL$ $V = m/(rho) = "2916.5 g"/"3.6 g/mL" = "810. mL"$

As you can see, the only difference between these two methods is that in the first one you multiply by $\frac{1}{ρ}$ $1/(rho)$ , while in the second one you divide by $ρ$ $rho$ , which is of course equivalent from a mathematical standpoint.

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Question #76f75

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