# 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".

Sep 15, 2017

$c = \text{9.48 mol/L}$

Well, the solution is aqueous, so it contains ${\text{KNO}}_{3}$ 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:

${\chi}_{K N {O}_{3}} = {n}_{K N {O}_{3}} / \left({n}_{K N {O}_{3}} + {n}_{{H}_{2} O}\right) = 0.195$

with ${\chi}_{K N {O}_{3}} + {\chi}_{{H}_{2} O} = 1$.

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 $\text{1 mol}$ for the total mols. This gives us:

$0.195 \cancel{{\text{mols KNO"_3) xx ("39.098 g K" + "14.007 g N" + 3 xx "15.999 g NO"_3)/cancel("1 mol KNO}}_{3}}$

$= {\text{13.475 g KNO}}_{3}$

$0.805 \cancel{\text{mols H"_2"O") xx ("18.015 g H"_2"O")/cancel("1 mol H"_2"O}}$

$= \text{14.502 g H"_2"O}$

And so the total mass of solution is:

$\text{13.475 g KNO"_3 + "14.502 g H"_2"O" = "27.977 g soln}$

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

$27.977 \cancel{\text{g soln" xx "1 mL soln"/(1.36 cancel"g") = ul"20.572 mL soln}}$

$=$ $\underline{\text{0.020572 L soln}}$

Lastly, we know the mols of ${\text{KNO}}_{3}$ already, so using the definition of molarity, we get:

$\textcolor{b l u e}{c} = \text{mols KNO"_3/"L soln}$

$= \left(\text{0.195 mols KNO"_3)/("0.020572 L soln}\right)$

$=$ $\underline{\textcolor{b l u e}{\text{9.48 mol/L}}}$

What if we had ${\text{0.390 mols KNO}}_{3}$ and $\text{1.61 mols H"_2"O}$? Would we have to reconsider our molarity? What does this illustrate about molarity, then? Is it intensive or extensive?