What is the percentage concentration by mass of a 16*mol*L^-1 solution of nitric acid, for which density is 1.42*g*mL^-1?

Oct 5, 2015

Around 70% concentration, mass of acid/mass of solution.

Explanation:

A $16$ $M$ concentration refers to the concentration of the solution. That is the nitric acid solution has a concentration of $16$ $m o l$ ${L}^{-} 1$ of solution . We can work out percentages ($\frac{w}{w}$ xx100%) provided that we know the density of the solution, which we do because you have kindly included it in your question. So $1$ $L$ of solution has a mass of $1420$ $g$. In the solution there are $16$ $m o l$ $H N {O}_{3}$, i.e. $16$ $m o l$ $\times$ $63.0$ $g$ $m o {l}^{-} 1$ $=$ $1008.2$ $g$.

So now we calculate the quotient, 1008.2 g $\times$ 1/(1420 g) $\times$ 100% $=$ ?? %. And this gives us our percentage concentration: (mass of solute/mass of solution) $\times$ 100%

Oct 5, 2015

Your stock solution is 71%"w/w" nitric acid.

Explanation:

What you need to do here is pick a sample of this stock nitric acid solution and use its molarity to find out how much nitric acid it would contain.

Then use its density to determine athe sample's mass.

So, to make calculations easier, pick a $\text{1.00-L}$ sample of the stock solution. Since molarity is defined as moles of solute, in your case nitric cid, divided by liters of solution, you get that

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{H N {O}_{3}} = \text{16 M" * "1.00 L" = "16 moles}$

Use nitric acid's molar mass to help you find how many grams of acid would contain this many moles

16color(red)(cancel(color(black)("moles HNO"""_3))) * "63.013 g"/(1color(red)(cancel(color(black)("mole HNO"""_3)))) = "1008.2 g HNO"""_3

What would be the mass of the sample? Use its known density!

1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.42 g"/(1color(red)(cancel(color(black)("mL")))) = "1420 g"

The solution's percent concentration by mass, or $\text{%w/w}$, is defined as

$\text{%w/w" = "mass of solute"/"mass of solution} \times 100$

In your case, you would have

"%w/w" = (1008.2color(red)(cancel(color(black)("g"))))/(1420color(red)(cancel(color(black)("g")))) xx 100 = color(green)(71%)

SIDE NOTE You can redo the calculations using any sample of the stock solution, the result will always come out the same.