# Find number grams of the solute in a dissolution given its density and Molarity

Have a dissolution of $\ce{HNO_3}$ in water. The density is $1.42\ce{g/mL}$ with a concentration of $\ce{16M}$. What is the concentration in % mass/mass?

Alright. So the % mass/mass is expressed as

$$\frac{\text{grams of } \ce{HNO_3}}{\text{grams of the dissolution}}\cdot100$$

So I need to find the grams of $\ce{HNO_3}$ first. They tell me that the Molarity is

$$16 = \frac{\text{moles of } \ce{HNO_3}}{\text{liters of dissolution}}$$

$\ce{HNO_3}$ has a molar mass of $63$. So to find out how many moles of $\ce{HNO_3}$ are in the dissolution I need to divide the grams of $\ce{HNO_3}$ by $63$... But how am I supposed to find out the grams of $\ce{HNO_3}$?

I was told that the numerator is $63$. But... $63$ is the molar mass of $\ce{HNO_3}$, not the number of moles of $\ce{HNO_3}$ in the dissolution.

Anyway, let's suppose that it is indeed $63$. Therefore

$$16 = \frac{63}{\text{liters of dissolution}} \implies \text{liters of dissolution} = 3.9$$

So there are $3.9$ liters of dissolution? I'm not sure - I feel like I got this all wrong. How do you know there are $63$ moles of $\ce{HNO_3}$?

• Ok, it all makes sense, but what I don't understand is why are you picking $1L$ of solution? Why one and not, say, five liters? Where/how do you get that number? In other words, how do you even know there is one liter. – Voldemort Jul 15 '15 at 6:23
• Well I meant the Molarity part. The Molarity formula in your case is $16 = \frac{63}{1}$, but if you change that $1$ by a $2.8$ instead of getting $1008$ grams you would get 360 grams of $\ce{HNO_3}$. – Voldemort Jul 15 '15 at 6:35
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