Question 10436

Apr 15, 2016

${\text{0.452 mol L}}^{- 1}$

Explanation:

In order to be able to calculate this solution's molarity, you must first determine how many moles of ethanol, $\text{C"_2"H"_5"OH}$, are present in that $\text{1.77-g}$ sample.

To do that, use ethanol's molar mass, which is equal to ${\text{46.07 g mol}}^{- 1}$. As you know, a compound's molar mass tells you the mass of one mole of that compound.

In this case, one mole of ethanol has a mass of $\text{46.07 g}$, which means that your sample will contain

1.77 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_5"OH")/(46.07color(red)(cancel(color(black)("g")))) = "0.03842 moles C"_2"H"_5"OH"

Now, a solution's molarity tells you how many moles of solute you get per liter of solution. This means that you can find a solution's molarity by finding out how many moles of solute you get in one liter of solution.

Your solution is said to have a volume of $\text{85.0 mL}$. Convert this to liters by using

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

85.0 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 85.0 * 10^(-3)"L"

So, the number of moles of solute present in one liter of this solution will be

1 color(red)(cancel(color(black)("L solution"))) * ("0.03842 moles C"_2"H"_5"OH")/(85.0 * 10^(-3)color(red)(cancel(color(black)("L solution")))) = "0.452 moles C"_2"H"_5"OH"

Since one liter of solution contains $0.452$ moles of ethanol, it follows that the solution's molarity will be equal to

"molarity" = c = color(green)(|bar(ul(color(white)(a/a)"0.452 mol L"^(-1)color(white)(a/a)|)))#

The answer is rounded to three sig figs.