# Question #b1009

##### 1 Answer
Jul 14, 2017

$\text{Molarity"="Moles of acid"/"Volume of soltion.}$

And by the way you add acid to water, and never water to acid..

#### Explanation:

We have a molar quantity of ..........................

$7.2 \cdot m o l \cdot {L}^{-} 1 \times 455.2 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 = 3.28 \cdot m o l$ with respect to $H C l$.

And thus.............. $\text{Molarity} = \frac{7.2 \cdot m o l \cdot {L}^{-} 1 \times 455.2 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}{1270.0 \cdot L}$

$= 2.258 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$