Question #c5f4d

Molarity is simply a measure of a solution's concentration in terms of how many moles of solute it contains per liter of solution.

This means that every time a problem wants you to find a solution's molarity, you must look for two things

• the number of moles of solute it contains
• the volume of the solution

In your case, a solution is said to contain #"125 g"# of methanol, #"CH"_3"OH"#, and have a total volume of #"0.25 L"#.

You already have the volume of the solution, so all you need to determine is how many moles of methanol are present in that many grams.

To do that, use methanol's molar mass as a conversion factor

#125 color(red)(cancel(color(black)("g"))) * overbrace(("1 mole CH"_3"OH")/(32.042color(red)(cancel(color(black)("g")))))^(color(purple)("the molar mass of CH"_3"OH")) = "3.901 moles CH"_3"OH"#

So, you know that #"0.25 L"# of solution contains #3.901# moles of methanol, your solute. Since molarity expresses the number of moles per liter of solution, use this as a conversion factor to find

#1 color(red)(cancel(color(black)("L solution"))) * ("3.901 moles CH"_3"OH")/(0.25color(red)(cancel(color(black)("L solution")))) = "15.6 moles CH"_3"OH"#

So, if every liter of this solution contains #15.6# moles of methanol, you can say that it's molarity is equal to

#"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("16 g mol"^(-1))color(white)(a/a)|)))#

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution.