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

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")`

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"`