What is the molar concentration of methanol if the mol fraction of methanol is #0.040# at #4^@ "C"#?

2 Answers
Jun 9, 2017

About #"2.12 M"# if you can look up the density of methanol. About #"2.31 M"# if you cannot, which would be around #9%# error.


Recall that:

#"molarity" -= ("mol solute")/("L soln")#

Since we are given that the mol fraction of methanol is

#chi_(MeOH) = (n_(MeOH))/(n_(MeOH) + n_(H_2O)) = 0.040#,

we probably have to make an assumption here, as we are only given a relative quantity with the mol fraction.

Assume that you have #"1 L"# of water so that we can use the density of water (at #4^@ "C"#):

#cancel("1 L H"_2"O") xx (1000 cancel("g H"_2"O"))/cancel"L" xx ("1 mol H"_2"O")/(18.015 cancel("g H"_2"O"))#

#=# #"55.51 mols"#

(note that this also means that the molarity of pure water [at #4^@ "C"#] is #"55.51 M"#.)

Given we know the mols of water now, we can use the mol fraction to solve for the mols of methanol:

#0.040 = n_(MeOH)/(n_(MeOH) + 55.51)#

#=> 0.040n_(MeOH) + 0.040(55.51) = n_(MeOH)#

#=> 0.960n_(MeOH) = 2.220#

#=> n_(MeOH) = "2.313 mols"#

This then gives us a molarity of:

#"2.313 mols MeOH"/V_(sol n)#

#"2.313 mols MeOH"/(V_(MeOH) + V_(H_2O))#

#= "2.313 mols MeOH"/(n_(MeOH)barV_(MeOH) + V_(H_2O))#

where #barV# is the molar volume in #"L/mol"#.

In a situation where you cannot look up the density of methanol, we would have had to assume that #V_(sol n) ~~ V_(H_2O)# for a sufficiently dilute solution, and we would have gotten #color(red)("2.313 M")#. That is, however, not entirely accurate.

We'll simply look up the density of methanol, which is #~~# #"806 g/L"# (by interpolation at #"277.15 K"#). So, we have:

#barV_(MeOH) = "1 L"/(806 cancel"g MeOH") xx (32.04 cancel"g MeOH")/("mol") = "0.0398 L/mol"#

This means the volume of methanol in the solution is:

#V_(MeOH) = n_(MeOH)barV_(MeOH )#

#= "2.313 mols" xx "0.0398 L/mol" = "0.0919 L"#

Assuming additivity of volumes of water and methanol:

#V_(sol n) ~~ V_(MeOH) + V_(H_2O) ~~ "1.0919 L soln"#

So, the molarity is:

#color(blue)(["MeOH"(aq)]) ~~ "2.313 mols"/"1.0919 L" = color(blue)("2.12 M")#

Oct 2, 2017

Answer:

The molarity is 2.1 mol/L.

Explanation:

Assume that we have a solution containing 0.040 mol methanol and 0.960 mol water (#chi_text(MeOH) = 0.040#).

#"Volume of MeOH" = 0.040 color(red)(cancel(color(black)("mol MeOH"))) × (32.04 color(red)(cancel(color(black)("g MeOH"))))/(1 color(red)(cancel(color(black)("mol MeOH")))) × "1 mL MeOH"/(0.792 color(red)(cancel(color(black)("g MeOH")))) = 1.62 "mL MeOH"#

#"Volume of H"_2"O" = 0.960 color(red)(cancel(color(black)("mol H"_2"O"))) × (18.02 color(red)(cancel(color(black)("g H"_2"O"))))/(1 color(red)(cancel(color(black)("mol H"_2"O")))) × ("1 mL H"_2"O")/(1.0 color(red)(cancel(color(black)("g H"_2"O")))) = "17.3 mL H"_2"O"#

If there is no volume change on mixing,

#"Total volume" = "(1.62 + 17.3) mL" = "18.9 mL"#

Then,

#"Molarity" = "moles of methanol"/"litres of solution" = "0.040 mol"/"0.0189 L" = "2.1 mol/L"#