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".

1 Answer
Sep 15, 2017

c = "9.48 mol/L"


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?