Question

Question #658b0

Answer 1

`"14 mol L"^(-1)`

Explanation:

Assuming that you can dissolve `3.4` moles of this unknown solute in enough water to make `"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

`"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

`"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` `"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 `10^3color(white)(.)"mL" = "1 L"` of solution, you can say that the molarity of the solution is

`color(darkgreen)(ul(color(black)("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.