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?

Well, a molarity of 16 means that there are 16 moles of solute for every liter of solution.

Based on density, if we have 1000 milliliters, we have 1420 grams of solution.

To calculate the molality I need:

$$\frac{16}{\text{kilograms of solvent}}$$

Since we have 16 moles, and the molar mass of $\ce{HNO3}$ is 63, we know that we have 1008 grams of solute, meaning there are 412 grams of solvent.

So I calculate the molality to be 0.038.

This answer is wrong because it is not among the options. What did I do wrong?

  • $\begingroup$ This appears to be a duplicate of multiple questions. $\endgroup$ Nov 6, 2015 at 3:00
  • $\begingroup$ 1 molar = 1 mole/(liter of solution) $\endgroup$
    – MaxW
    Nov 6, 2015 at 3:10
  • $\begingroup$ @Max Your definition of molal is incorrect. It should be amount of substance of solute per mass of solvent. $\endgroup$ Nov 6, 2015 at 7:24
  • $\begingroup$ 1 molal = 1 mole / (kg of water) $\endgroup$
    – MaxW
    Nov 6, 2015 at 7:40

2 Answers 2


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 $M(\ce{HNO3})=63~\mathrm{g\,mol^{-1}}$.


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.)


The concentration $c$, as fraction of molecular amount $n$ per volume $V$ takes into account both the volume of the solvent plus the volume(s) of the solute(s). In other words, the total volume of the whole solution. Eventually, to determine the molalilty, you divide the number of moles of the solute per kilogram solvent.


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