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

Answer:

The correct answer is "11.44 ml".

Explanation:

Molarity,

= 0.2 M

Density,

= 1.05 g/ml

Volume,

= 1 L

As we know,

⇒  $Molarity=\frac{No. \ of \ moles }{Volume \ of \ solution}$

or,

⇒  $No. \ of \ moles=Molarity\times Volume$

On putting the values, we get

⇒                         $=0.2\times 1$

⇒                         $=0.2 \ moles$

Now,

⇒  $No. \ of \ moles=\frac{Mass \ taken}{Molecular \ mass}$

or,

⇒  $Mass \ taken=No. \ of \ moles\times Molecular \ mass$

⇒                      $=0.2\times 60.05$

⇒                      $=12.01 \ gram$

hence,

⇒  $Density= \frac{Mass }{Volume}$

or,

⇒  $Volume=\frac{Mass}{Density}$

⇒                $=\frac{12.01}{1.05}$

⇒                $=11.44 \ ml$