# A solution is prepared from 8 grams of acetic acid (CH_2CO_2H) and 725 grams of water. What is the molarity of this solution?

Aug 15, 2016

$\text{Molarity}$ $=$ $\text{Moles of solute"/"Volume of solution}$

#### Explanation:

$\text{Molarity}$ $=$ $\frac{\frac{8 \cdot g}{60.05 \cdot g \cdot m o {l}^{-} 1}}{0.725 \cdot L}$ $\cong$ $0.18 \cdot m o l \cdot {L}^{-} 1$.

Note that I have assumed here that the volume of the acid in the given mass of water is not going to vary substantially from $0.725 \cdot L$, which was the standard volume of the mass of water used to dilute it. This is a reasonable assumption.

If we liked, we could calculate the $\text{molality}$ of the solution:

$\text{Moles of solute"/"kg of solvent}$.