Question

You must make 1 L of 0.2 M acetic acid (CH3COOH). All you have available is concentrated glacial acetic acid (assay value, 98%; specific gravity, 1.05 g/mL). It will take _________ milliliters of acetic acid to make this solution. Assume a gram molecular weight of 60.05 grams.

Answer 1

The correct answer is "11.44 ml".

Step-by-step explanation:

Molarity,

= 0.2 M

Density,

= 1.05 g/ml

Volume,

= 1 L

As we know,

⇒
`Molarity=(No. \ of \ moles )/(Volume \ of \ solution)`

or,

⇒
`No. \ of \ moles=Molarity* Volume`

On putting the values, we get

⇒
`=0.2* 1`

⇒
`=0.2 \ moles`

Now,

⇒
`No. \ of \ moles=(Mass \ taken)/(Molecular \ mass)`

or,

⇒
`Mass \ taken=No. \ of \ moles* Molecular \ mass`

⇒
`=0.2* 60.05`

⇒
`=12.01 \ gram`

hence,

⇒
`Density= (Mass )/(Volume)`

or,

⇒
`Volume=(Mass)/(Density)`

⇒
`=(12.01)/(1.05)`

⇒
`=11.44 \ ml`