Question

Question #150ce

Answer 1

You'd need 27 mL of pure methanol to make that solution.

Explanation:

So, you know that you have a stock bottle of liquid methanol. The methanol has a density of 0.793 kg/L, which means that a 1-L bottle will contain 0.793 kg of methanol.

To make the calculations easier, convert the density from kg per liter to g per liter

`0.793cancel("kg")/"L" * "1000 g"/(1cancel("kg")) = "793 g/L"`

Now, you know that your target solution has a molarity of 0.25 M and a volume of 2.5 L. This means that you can determine how many moles of methanol the target solution must contain

`C = n/V => n = C * V`

`n_"methanol" = "0.25 M" * "2.5 L" = "0.625 moles"`

To determine how many grams of methanol would contain this many moles, use the compound's molar mass

`0.625cancel("moles") * "34.04 g"/(1cancel("mole")) = "21.275 g"`

This is how much methanol your target solution must contain. To determine what volume of the pure methanol solution you need to get this many grams, use the given density

`21.275cancel("g") * "1 L"/(793cancel("g")) = "0.0268 L"`

Rounded to two sig figs, the number of sig figs you gave for the target solution's volume and molarity, and expressed in mililiters, the answer will be

`0.0268cancel("L") * "1000 mL"/(1cancel("L")) = 26.8 = color(green)("27 mL")`

So, if you take 27 mL of pure methanol, and add enough water to make a total volume of 2.5 L, you'll end up with a 0.25-M solution.