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

Jun 9, 2017

About $\text{2.12 M}$ if you can look up the density of methanol. About $\text{2.31 M}$ 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

${\chi}_{M e O H} = \frac{{n}_{M e O H}}{{n}_{M e O H} + {n}_{{H}_{2} O}} = 0.040$,

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

Assume that you have $\text{1 L}$ of water so that we can use the density of water (at ${4}^{\circ} \text{C}$):

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"))

$=$ $\text{55.51 mols}$

(note that this also means that the molarity of pure water [at ${4}^{\circ} \text{C}$] is $\text{55.51 M}$.)

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

$0.040 = {n}_{M e O H} / \left({n}_{M e O H} + 55.51\right)$

$\implies 0.040 {n}_{M e O H} + 0.040 \left(55.51\right) = {n}_{M e O H}$

$\implies 0.960 {n}_{M e O H} = 2.220$

$\implies {n}_{M e O H} = \text{2.313 mols}$

This then gives us a molarity of:

$\frac{\text{2.313 mols MeOH}}{V} _ \left(s o l n\right)$

$\frac{\text{2.313 mols MeOH}}{{V}_{M e O H} + {V}_{{H}_{2} O}}$

$= \frac{\text{2.313 mols MeOH}}{{n}_{M e O H} {\overline{V}}_{M e O H} + {V}_{{H}_{2} O}}$

where $\overline{V}$ is the molar volume in $\text{L/mol}$.

In a situation where you cannot look up the density of methanol, we would have had to assume that ${V}_{s o l n} \approx {V}_{{H}_{2} O}$ for a sufficiently dilute solution, and we would have gotten $\textcolor{red}{\text{2.313 M}}$. That is, however, not entirely accurate.

We'll simply look up the density of methanol, which is $\approx$ $\text{806 g/L}$ (by interpolation at $\text{277.15 K}$). So, we have:

${\overline{V}}_{M e O H} = \text{1 L"/(806 cancel"g MeOH") xx (32.04 cancel"g MeOH")/("mol") = "0.0398 L/mol}$

This means the volume of methanol in the solution is:

${V}_{M e O H} = {n}_{M e O H} {\overline{V}}_{M e O H}$

$= \text{2.313 mols" xx "0.0398 L/mol" = "0.0919 L}$

Assuming additivity of volumes of water and methanol:

${V}_{s o l n} \approx {V}_{M e O H} + {V}_{{H}_{2} O} \approx \text{1.0919 L soln}$

So, the molarity is:

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

Oct 2, 2017

The molarity is 2.1 mol/L.

#### Explanation:

Assume that we have a solution containing 0.040 mol methanol and 0.960 mol water (${\chi}_{\textrm{M e O H}} = 0.040$).

$\text{Volume of MeOH" = 0.040 color(red)(cancel(color(black)("mol MeOH"))) × (32.04 color(red)(cancel(color(black)("g MeOH"))))/(1 color(red)(cancel(color(black)("mol MeOH")))) × "1 mL MeOH"/(0.792 color(red)(cancel(color(black)("g MeOH")))) = 1.62 "mL MeOH}$

$\text{Volume of H"_2"O" = 0.960 color(red)(cancel(color(black)("mol H"_2"O"))) × (18.02 color(red)(cancel(color(black)("g H"_2"O"))))/(1 color(red)(cancel(color(black)("mol H"_2"O")))) × ("1 mL H"_2"O")/(1.0 color(red)(cancel(color(black)("g H"_2"O")))) = "17.3 mL H"_2"O}$

If there is no volume change on mixing,

$\text{Total volume" = "(1.62 + 17.3) mL" = "18.9 mL}$

Then,

$\text{Molarity" = "moles of methanol"/"litres of solution" = "0.040 mol"/"0.0189 L" = "2.1 mol/L}$