# Question 658b0

Aug 26, 2017

${\text{14 mol L}}^{- 1}$

#### Explanation:

Assuming that you can dissolve $3.4$ moles of this unknown solute in enough water to make $\text{245 mL}$ of solution at room temperature, you can calculate the molarity of this solution by figuring out how many moles of solute would be present in

$\text{1 L" = 10^3color(white)(.)"mL}$

of this solution.

Now, solutions are homogeneous mixtures, which implies that they same composition throughout.

This will let you use the known composition of the solution

$\text{3.4 moles solute " -> " 245 mL solution}$

as a conversion factor to help you find the number of moles of solute that would be present in ${10}^{3}$ $\text{mL}$ of this solution.

You're going from volume of solution to moles, so set up the conversion factor as

"3.4 moles solute"/"245 mL solution" color(white)(aa)color(white)((color(blue)( larr " what you need goes on top")aaaaaa)/color(blue)(larr " what you have goes on the bottom"))

You should end up with

10^3 color(red)(cancel(color(black)("mL solution"))) * "3.4 moles solute"/(245 color(red)(cancel(color(black)("mL solution")))) = "13.88 moles solute"#

Since this represents the number of moles of solute present in ${10}^{3} \textcolor{w h i t e}{.} \text{mL" = "1 L}$ of solution, you can say that the molarity of the solution is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity" = "14 mol L}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of solute present in your sample.