Question

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

Answer 1

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

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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).