Question

Question #df175

Answer 1

Here's what I got.

Explanation:

Your strategy here will be to use the given volume and the densities of the three substances to determine how many grams of each you're mixing.

Once you know that, you can use their molar masses to convert the grams to moles.

So, you know that you're mixing `"20 cm"^3` of ethyl alcohol, `"CH"_3"CH"_2"OH"`, methyl alcohol, `"CH"_3"OH"`, and water, `"H"_2"O"`.

As you know, the density of a substance tells you what mass of said substance you get per unit of volume. In this case, the unit of volume is `"1 cm"^3`. This means that you will have

`"CH"_3"CH"_2"OH: " rho = "0.79 g cm"^(-3) -> "0.79 g"` for every `"1 cm"^(3)`

`color(white)("CH"_2)"CH"_3"OH: " rho = "0.81 g cm"^(-3) -> "0.81 g"` for every `"1 cm"^3`

`color(white)("CH"_3"CH")"H"_2"O: " rho = color(white)(a)"1.0 g cm"^(-3) -> color(white)(a)"1.0 g"` for every `"1 cm"^3`

This means that your `"20 cm"^3` samples will contain

`20 color(red)(cancel(color(black)("cm"^3))) * "0.79 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "15.8 g CH"_3"CH"_2"OH"`

`20 color(red)(cancel(color(black)("cm"^3))) * "0.81 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "16.2 g CH"_3"OH"`

`20 color(red)(cancel(color(black)("cm"^3))) * "1.0 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "20 g H"_2"O"`

Now use the molar masses of the three compounds to convert the grams to moles

`15.8 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"CH"_2"OH")/(46.07 color(red)(cancel(color(black)("g")))) = "0.343 moles CH"_3"CH"_2"OH"`

`16.2 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"OH")/(32.04color(red)(cancel(color(black)("g")))) = "0.506 moles CH"_3"OH"`

`20color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.11 moles H"_2"O"`

To get the mole fraction of each component of the mixture, calculate the total number of moles present

`n_"total" = 0.343 + 0.506 + 1.11 = "1.96 moles"`

The mole fraction of a component `i` of the system, `chi_i`, will be equal to

`color(blue)(|bar(ul(color(white)(a/a)chi_i = "number of moles of i"/"total number of moles"color(white)(a/a)|)))`

In your case, you will have

`chi_("CH"_3"CH"_2"OH") = (0.343 color(red)(cancel(color(black)("moles"))))/(1.96color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.175)color(white)(a/a)|)))`

`chi_("CH"_3"OH") = (0.506 color(red)(cancel(color(black)("moles"))))/(1.96color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.258)color(white)(a/a)|)))`

`chi_("H"_2"O") = (1.11 color(red)(cancel(color(black)("moles"))))/(1.96color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.566)color(white)(a/a)|)))`

I'll leave the answers rounded to three sig figs, but keep in mind that you only have one sig fig for the volume of the samples.