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
c = "9.48 mol/L"
Well, the solution is aqueous, so it contains
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
It is convenient to choose to have
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
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