Question

How many milliliters of pure liquid methanol (ch3oh, mw = 32.04 g/mol) are needed to prepare 500.0 ml of an aqueous solution of 12.0 g/l methanol? the density of pure liquid methanol is 0.791 g/ml?

Answer 1

Approximately 7589.9 mL of pure liquid methanol are needed to prepare 500.0 mL of a 12.0 g/L methanol solution.

Step-by-step explanation:

To calculate the volume of liquid methanol needed to prepare the aqueous solution, we can use the equation:

mass of methanol (g) = volume of methanol (mL) * density of methanol (g/mL)

First, we need to convert the mass of methanol in the solution to grams:

mass of methanol (g) = 500.0 mL * 12.0 g/L = 6000.0 g

Next, we can use the density of methanol to find the volume of pure liquid methanol needed:

volume of methanol (mL) = mass of methanol (g) / density of methanol (g/mL) = 6000.0 g / 0.791 g/mL = 7589.9 mL

Therefore, approximately 7589.9 mL of pure liquid methanol are needed to prepare 500.0 mL of the 12.0 g/L methanol solution.

Answer 2

Answer:- 7.59 mL.

Solution:- We want to make 500.0 mL of an aqueous solution of methanol whose density is
`12.0(g)/(L)`. This has to be made from a given pure liquid methanol with density
`0.791(g)/(mL)` . It asks to calculate the volume of pure liquid methanol.

We know that, mass = volume*density

Let's calculate the mass of aqueous solution from it's given volume and density. need to convert volume unit from mL to L as the density is given in grams per liter:

mass of aqueous methanol solution =
`500.0mL((1L)/(1000mL))((12.0g)/(L))`

= 6 g

From above calculations, we need to use 6 g of pure liquid methanol. It's density is known, so we could calculate it's volume as:

`volume=(mass)/(density)`

Let's plug in the values in it:

`volume=6g((1mL)/(0.791g))`

= 7.59 mL

Hence, we need to take 7.59 mL of pure liquid methanol.