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"#.

#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?