# What volume of 84% HNO_3 "w/w", rho=1.54*g*mL^-1, is required to make a 1.0*L of a 1*mol*L^-1 solution of nitric acid?

Sep 3, 2017

Approx. $25 \cdot m L$

#### Explanation:

We use the relationship, $\text{concentration"="moles of solute"/"volume of solution}$

And thus we need $1 \cdot L \times 0.50 \cdot m o l \cdot {L}^{-} 1 = 0.50 \cdot m o l$ with respect to nitric acid. And this represents a mass of $0.50 \cdot m o l \times 63.01 \cdot g \cdot m o {l}^{-} 1 = 31.5 \cdot g$.

We gots 84% $H N {O}_{3}$, $\text{w/w}$............

And thus we need a mass of solution......

$\frac{31.5 \cdot g}{0.84} = 37.5 \cdot g$.......

And given the density......$\rho = 1.54 \cdot g \cdot m {L}^{-} 1$, we need....

$\frac{37.5 \cdot g}{1.54 \cdot g \cdot m {L}^{-} 1} = 24.4 \cdot m L .$

Nitric acid of this concentration is called in the trade $\text{red fuming nitric acid}$, and its density owes to the solubility of $N {O}_{2}$.