If a solution of $\ce{HNO3}$ has a molarity of 16 and a density of
$1.42\rm{g/mL}$, what is the molality of the solution?
Molarity is an old word for amount concentration and it is defined by the IUPAC Goldbook as
Amount of a constituent divided by the volume of the mixture. Also called amount-of-substance concentration, substance concentration (in clinical chemistry) and in older literature molarity. For entities B it is often denoted by B. The common unit is mole per cubic decimetre ($\mathrm{mol\,dm^{−3}}$) or mole per litre ($\mathrm{mol\,L^{−1}}$) sometimes denoted by $\mathrm{M}$.
In mathematical terms $$c(\ce{HNO3(aq)})=\frac{n(\ce{HNO3})}{V(\ce{HNO3(aq)}}.$$
In this case the concentration is presumably given as $c(\ce{HNO3(aq)})=16~\mathrm{mol\,L^{-1}}$
The molality, $b$, is defined by the IUPAC Goldbook as
Amount of entities of a solute divided by the mass of the solvent.
In mathematical terms this means
$$b(\ce{HNO3(aq)})=\frac{n(\ce{HNO3})}{m(\ce{H2O})}.$$
The (mass) density, $\rho$, is defined by the IUPAC Goldbook as
Mass of a sample or body divided by its volume.
In mathematical terms this means
$$\rho(\ce{HNO3(aq)})=\frac{m(\ce{HNO3(aq)})}{V(\ce{HNO3(aq)})}.$$
In this case the density is given as $\rho(\ce{HNO3(aq)})=1.42~\mathrm{g\,mL^{-1}}.$
Now you can go ahead and derive the formula for conversion all by yourself, which I recommend, but you can also check if someone has already done that. I chose to be lazy today, so I checked Wikipedia:
\begin{align}
c &= \frac{\rho b}{1 + bM} &
b &= \frac{c}{\rho - cM}
\end{align}
Now you only have to figure out the molar mass of nitric acid, and you can calculate the result.
I assume $M(\ce{HNO3})=63~\mathrm{g\,mol^{-1}}$.
Solution:
You should find $b(\ce{HNO3(aq)})=38.8~\mathrm{mol\,kg^{-1}}$.
So in principle you have done quite good. If you would have used proper notation and included all the units in every step, you would maybe have noticed the mistake all on your own.
(Hover over the hidden fields to reveal their content.)
