# Question 150ce

Jun 30, 2015

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

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{\text{methanol" = "0.25 M" * "2.5 L" = "0.625 moles}}$

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

$0.0268 \cancel{\text{L") * "1000 mL"/(1cancel("L")) = 26.8 = color(green)("27 mL}}$

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.