Question

What is the molarity for a solution with a mol fraction of `0.195` for the potassium nitrate solute? Its density is `"1.36 g/mL"`.

Answer 1

`c = "9.48 mol/L"`

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Well, the solution is aqueous, so it contains `"KNO"_3` and water (assume for the moment that we are considering it as a whole instead of its dissociated parts).

Then, the mol fraction is given by:

`chi_(KNO_3) = n_(KNO_3)/(n_(KNO_3) + n_(H_2O)) = 0.195`

with `chi_(KNO_3) + chi_(H_2O) = 1`.

Therefore, having a two-component mixture, the mol fraction of water is `0.805`. Since all mol fractions are relative quantities, we need to assume an actual quantity of each substance.

It is convenient to choose to have `"1 mol"` for the total mols. This gives us:

`0.195 cancel("mols KNO"_3) xx ("39.098 g K" + "14.007 g N" + 3 xx "15.999 g NO"_3)/cancel("1 mol KNO"_3)`

`= "13.475 g KNO"_3`

`0.805 cancel("mols H"_2"O") xx ("18.015 g H"_2"O")/cancel("1 mol H"_2"O")`

`= "14.502 g H"_2"O"`

And so the total mass of solution is:

`"13.475 g KNO"_3 + "14.502 g H"_2"O" = "27.977 g soln"`

Using its density, we can then convert to its volume:

`27.977 cancel"g soln" xx "1 mL soln"/(1.36 cancel"g") = ul"20.572 mL soln"`

`=` `ul"0.020572 L soln"`

Lastly, we know the mols of `"KNO"_3` already, so using the definition of molarity, we get:

`color(blue)(c) = "mols KNO"_3/"L soln"`

`= ("0.195 mols KNO"_3)/("0.020572 L soln")`

`=` `ulcolor(blue)("9.48 mol/L")`

What if we had `"0.390 mols KNO"_3` and `"1.61 mols H"_2"O"`? Would we have to reconsider our molarity? What does this illustrate about molarity, then? Is it intensive or extensive?