# What is the molarity of 53%"w/w" "HNO"_3?

Feb 4, 2018

#### Answer:

$M = 0.53$

#### Explanation:

The concentration is the molarity, except that molarity is in decimal form.

Molarity, $c$, is defined as $c = \frac{n}{V}$, where $n$ is the moles of the solute and $V$ the volume of solution.

Here, the molarity will be $0.53$.

Mar 1, 2018

I'm getting a bit over $\text{11 M}$.

You have

"53 g HNO"_3/("100 g solution")

And at ${20}^{\circ} \text{C}$, it has density of $\text{1.3278 g/mL}$. Therefore, this has a volume of:

$100 \cancel{\text{g solution" xx "1 mL"/(1.3278 cancel"g soln") = "75.31 mL}}$

$=$ $\text{0.07531 L}$

As a result, this has a molarity of:

color(blue)(["HNO"_3]_(53%)) = (53 cancel"g" xx "1 mol"/(63.0119 cancel("g HNO"_3)))/("0.07531 L")

$\approx$ $\textcolor{b l u e}{\text{11.17 M}}$