Calculate the volume in milliliters of 2.35 M potassium hydroxide that contains 13.5 g of solute?

#"102.4 mL"# of the solution contains #"13.5 g"# of #"KOH"#, having a concentration of #"2.35 mols KOH"/"L solution"#.

Once you work out the keywords, the math isn't too bad. First, let's define some terms:

• #"Molarity" = "mols solute"/"L solution" = "M" = "mols"/"L"#
• #"Solute" = "KOH" -= "potassium hydroxide"#
• #"Solvent" = "H"_2"O"# (water)

You are asked to get the #"mL"# (milliliters) of solution that contains #"13.5 g"# of #"KOH"#, and you are given the molarity, #"2.35 mols/L"#.

So, this is asking you to use the molarity and do a unit conversion. Since you want the volume for your answer, you would start with the orientation of the fraction so that the volume starts on top.

#"L"/(2.35 cancel"mols") xx cancel"1 mol KOH"/(56.1056 cancel"g KOH") xx 13.5 cancel"g KOH"#

#=# #"0.1024 L solution"#

But since you were asked for this in #"mL"#, there is one more step. There are 1000 millithings in one thing. So:

#color(blue)(V_(sol n)) = 0.1024 cancel"L" xx "1000 mL"/cancel"L"#

#=# #color(blue)("102.4 mL solution")#

For perspective, in real life, a #"mL"# (a cubic centimeter of water) is 1000 times smaller than #"1 L"#. If we have #"102.4 mL"#, that's a pretty ordinary amount in the laboratory (we normally have #"100.00 mL"#, #"200.00 mL"#, and #"250.00 mL"# volumetric flasks available).