Assuming that you can dissolve
#"1 L" = 10^3color(white)(.)"mL"#
of this solution.
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
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
#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.