# Question 47267

Jun 1, 2016

Here's what I got.

#### Explanation:

The key to this problem is the density of the solution.

The density of a substance essentially tells you the mass of one unit of volume of that substance. In your case, the density of the solution is said to be $2.5$ grams per milliliter, which means that one unit of volume will be $\text{1 mL}$.

So, the density of the solution tells you that every $\text{1 mL}$ of this solution has a mass of $\text{2.5 g}$.

You can thus use density as a conversion factor to help you convert between the mass of the solution and the volume it occupies.

Now, the problem tries to put you off track a little by telling you that

1000 mL of an extremely viscous solution has a density of $2.5$ grams/milliliter...

It doesn't matter how much of this solution you have, its density will always be the same.

This means that you can ignore the given $\text{1000 mL}$ altogether, that information is not important in this context.

A fluid ounce is approximately equal to

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 fl oz " = " 29.57 mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Use this conversion factor to go from fluid ounces to milliliters

2 color(red)(cancel(color(black)("fl oz"))) * "29.57 mL"/(1color(red)(cancel(color(black)("fl oz")))) = "59.14 mL"

Now, if $\text{1 mL}$ of this solution has a mass of $\text{2.5 g}$, it follows that $\text{59.14 mL}$ will have a mass of

59.14 color(red)(cancel(color(black)("mL"))) * overbrace("2.5 g"/(1color(red)(cancel(color(black)("mL")))))^(color(blue)("the density of the solution")) = "147.85 g"

You should round this off to one sig fig, the number of sig figs you have for the volume of the solution, i.e. $2$ fluid ounces, but I'll leave it rounded to two sig figs

"mass of 2 fl oz" = color(green)(|bar(ul(color(white)(a/a)color(black)("150 g")color(white)(a/a)|)))

Finally, to convert this to ounces, use the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 oz " = " 28.35 g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will have

150 color(red)(cancel(color(black)("g"))) * "1 oz"/(28.35color(red)(cancel(color(black)("g")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("5.3 oz")color(white)(a/a)|)))#