Question

A solution is prepared from 8 grams of acetic acid (CH2CO2H) and 725 grams of water. What is the molarity of this solution?

Answer 1

Answer:

0.19 m

Explanation:

Just took the test, this is correct :)

Answer 2

Hello!

First, we can assume the density of water to be 1 g/mL, so the volume of water would be 725 mL. The density of Acetic Acid (Pure) is 1,05 g/mL so 8 grams would represent 7,61 mL

Now we can apply the following conversion factor to calculate the molarity of the solution, using the molar mass of Acetic Acid:

$[CH_3CO_2H]=\frac{8 g_{CH_3CO_2H} }{725 mL + 7,61 mL}* \frac{1000 mL}{1 L}* \frac{1 mol_{CH_3CO_2H} }{60,05 g_{CH_3CO_2H} }$

$[CH_3CO_2H]=0,1818 M$

So, the concentration of acetic acid would be 0,1818 M

Have a nice day!