How many grams of nitric acid are present in 250.0 mL of 6.70 M acid solution?

Feb 10, 2016

Approx. 100 g.

Explanation:

$\text{Concentration} , m o l \cdot {L}^{-} 1$ $=$ $\text{Moles(n)"/"Volume of solution(L)}$.

Clearly, $n$, $\text{No. of moles}$ $=$ $\text{Concentration "xx" Volume}$.

So, $250 \times {10}^{- 3} \cancel{L} \times 6.70 \cdot m o l \cdot {\cancel{L}}^{-} 1$ $=$ ?? $m o l$ $H N {O}_{3}$.

And now, multiply that molar quantity by the molar mass of nitric acid, $63.01 \cdot g \cdot m o {l}^{-} 1$.

?? $m o l$ $\times$ $63.01 \cdot g \cdot m o {l}^{-} 1$ $=$ ?? $g$?