# What is the molarity of a 49.0*g mass of phosphoric acid in a 2.00*L volume of aqueous solution?

Sep 18, 2016

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

#### Explanation:

We get an answer with units $m o l \cdot {L}^{-} 1$.

$\text{Moles of phosphoric acid } \left({H}_{3} P {O}_{4}\right)$ $=$ $\frac{49.0 \cdot g}{97.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.5 \cdot m o l$.

$\text{Molarity}$ $=$ $\frac{0.50 \cdot m o l}{2000 \cdot c {m}^{3} \times {10}^{-} 3 c {m}^{3} \cdot {L}^{-} 1}$

$=$ $0.25 \cdot m o l \cdot {L}^{-} 1$ with respect to ${H}_{3} P {O}_{4}$.