Question

Question #46fd6

Answer 1

`"0.233 M"`

Explanation:

The idea here is that you can convert the solution's concentration from percent concentration by mass, or `%"m/m"`, to moles per liter, or molar, by picking a sample solution.

To make calculations easier, you can pick a `"100-g"` sample of the `"0.750 m/m"%` solution

By definition, a solution's percent concentration by mass is defined as the mass of the solute, in this case methanol, divided by the total mass of the solution, and multiplied by `100`.

`color(blue)("%m/m" = "mass of solute"/"mass of solution" xx 100)`

In your case, the `"100 g"` sample will contain `"0.750 g"` of methanol.

To get the number of moles of methanol, use the compound's molar mass

`0.750color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"OH")/(32.042 color(red)(cancel(color(black)("g")))) = "0.02341 moles CH"_3"OH"`

Now, to get the volume of the sample, you need to use its density. A `0.750%"m/m"` methanol solution will have a density of approximately `"0.997 g/mL"` at room temperature.

https://en.wikipedia.org/wiki/Methanol_%28data_page%29

This means that the sample will have a volume of

`100color(red)(cancel(color(black)("g"))) * "1 mL"/(0.997color(red)(cancel(color(black)("g")))) = "100.3 mL"`

Molarity is defined as moles of solute per liter of solution

`color(blue)(c = n/V)`

In your case, you would get - do not forget to convert the volume from mililiters to liters!

`c = "0.02341 moles"/(100.3 * 10^(-3)"L") = color(green)("0.233 M")`

The answer is rounded to three sig figs.