Question #10436

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

To do that, use ethanol's molar mass, which is equal to #"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 #"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 #"85.0 mL"#. Convert this to liters by using

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(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.