# A 30.00% (by mass) solution of HNO_3 in water has a density of 1.18 g/cm^3 at 20°C. What is the molarity of HNO_3 in the solution?

Apr 22, 2017

#### Answer:

We are given that "Mass of nitric acid"/"Mass of solution"xx100% $=$ 30.00%.

#### Explanation:

Now $\text{Concentration"="Moles of solute"/"Volume of solution}$.

How do we turn one t'other? Well, we need the density of the solution, which you have supplied, $\rho = 1.18 \cdot g \cdot c {m}^{-} 3$, and some tedious 'rithmetic................

So we work with a $1 \cdot L$ volume of solution, i.e. ${10}^{3} \cdot c {m}^{3}$.

"Moles of solute"=(30.00%xx10^3*cancel(cm^3)xx1.180*cancelg*cancel(cm^-3))/(63.01*cancelg*mol^-1)

$= 5.62 \cdot m o l$

But this molar quantity was dissolved in a $1 L$ volume of solvent.....

$\text{Concentration} = \frac{5.62 \cdot m o l}{1 \cdot L} = 5.62 \cdot m o l \cdot {L}^{-} 1$.

Now I happen to know that conc. nitric acid, which is commercially supplied as a 68%(w/w) solution, has a molar concentration of approx. $15 \cdot m o l \cdot {L}^{-} 1$, so the value we calculated is reasonably consistent. You should keep doing these types of problems; it's all too easy to get bewildered and make an error.