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.

The correct answer is "11.44 ml".

Explanation:

Molarity,

= 0.2 M

Density,

= 1.05 g/ml

Volume,

= 1 L

As we know,

⇒  

or,

⇒  

On putting the values, we get

⇒                        

⇒                        

Now,

⇒  

or,

⇒  

⇒                      

⇒                      

hence,

⇒  

or,

⇒  

⇒                

⇒                

the answer is c i believe

2b2t

explanation:

2b2t