Find the molality of a mixture formed by mixing $200cm^3$ of $HNO_3$ that has a $69\%$ of richness and density $1.41g/mL$, with $1L$ of the same acid with $1.2M$ and density $1.06g/mL$.

Attempt. The question asked the molarity too, and I think I've found it, but the issue comes with the molality. Pretty stuck here, I'm having troubles finding which is the solvent and which is the solute, is there any water going on here that it is assumed to be known? I know molality, let that be $m$, is defined as $m=\frac{n_s}{m_{solvent}}$, where $n_s$ and $m_{solvent}$ are the moles of the solute and the mass of the solvent respectively. But I want to be sure I know what the mixture is about. Are we mixing some mixture that has water and acid, with another mixture that has water and acid? Or is this just acid with acid, and the solvent is the acid with more proportion?

Continuation. Now that I know it is a mixture of 2 mixtures with both water and acid, let me post what I've tried:

I've used the letters to denote the first mixture, the letters with ' to denote the second mixture, and the letters with '' to denote the final mixure of both mixtures. $m''=\frac{n_s''}{m_{solvent}''}$, where $n_s''$ and $m_{solvent}''$ are the moles of the solute and the mass of the solvent of the new gas, denoted by the ''. I've found $n_s''$ by doing this: $$n''_s=n_s'+n_s$$, not sure if that thing above is true, but assuming it is, I've separated both mixtures with the information we are given about them: $$\text{Mixture }1: \begin{cases}m_{mixture}=V_{mixture}d_{mixture}\\ 0.69m_{mixture}=m_s\end{cases}$$ $$\text{Mixture }2: \begin{cases} 1.2=\frac{n_s'}{V_{mixture}'}\end{cases}$$, by simply just plugging the information we know and some unit conversions, with the equation above $$n''_s=n_s'+n_s$$ we get $$n_s''=1.2+0.69\cdot 0.2L\cdot 1410g/L\cdot \frac{1mole\ HNO_3}{63g\ HNO_3}\approx 4.289 moles$$ and we would have one part of the $m=\frac{n_s''}{m_{mixture}}$ done. Not sure on how to get the $m_{mixture}$ and not sure if what I've done is correct either.

  • 2
    $\begingroup$ Yes, we are mixing some mixture of water and acid with another mixture of water and acid. $\endgroup$ Jan 24 at 20:31
  • $\begingroup$ okay, let me add in then what I've tried so far to find the molality $\endgroup$
    – Xetrez
    Jan 24 at 20:33
  • $\begingroup$ done, i'll be back in half an hour if you answer. $\endgroup$
    – Xetrez
    Jan 24 at 20:47
  • 2
    $\begingroup$ Get rid of the word "gas". The rest sounds about right. Now, what's that with $m_{mixture}$? You don't need that; you need $m''_{solvent}$. Well, go on, find it. The solvent is water. The mass of both solutions can easily be found. Whatever of them is not acid, is necessarily water. $\endgroup$ Jan 24 at 20:58
  • $\begingroup$ yeah my bad I actually meant $m_{solvent}''$ $\endgroup$
    – Xetrez
    Jan 24 at 21:15

Ok, you correctly calculated the number of moles in the final solution with:

$$n''=n'+n$$ and

$$n''=\pu{1.2 M}\times \pu{1 L} + \dfrac{0.69\cdot \pu{0.2 L}\cdot \pu{1410 g/L}}{\pu{63.012 g/mol \ce{HNO_3}}} =\pu{4.2880 mol}$$

However I used 63.012 for the molecular mass of nitric acid instead of 63.

Now to calculat molarity you divide thus

$$\dfrac{4.2880}{\mathrm{liters\ of\ solution}} = \mathrm{molarity\ of\ \ce{HNO3}}$$

and to calculate molality you divide thus:

$$\dfrac{4.2880}{\mathrm{kg\ of\ solvent}} = \mathrm{molality\ of\ \ce{HNO3}}$$

At this point you have to assume that the volumes are additive, which isn't quite true..., so you end up with 1.2 L of solution. So for molarity:

$$\dfrac{4.2880}{\pu{1.2 L}} = \pu{3.57 M}\ \ce{HNO3}$$

Now the kg of solvent is a bit more convoluted. For the concentrated solution the mass of water is:

$$0.31\cdot \pu{0.2 L}\cdot \pu{1.410 kg/L} = \pu{0.08742 kg}$$

for the more dilute acid solution the mass of water is:

$$\pu{1.00 L} \cdot{(\pu{1.06 kg/L} - \pu{1.2 mol/L}\cdot \pu{0.063012 kg/mol} ) = \pu{0.98439 kg}}$$

so the total mass of water is

$$ \pu{0.08742 kg} + \pu{0.98439 kg} = \pu{1.07181 kg}$$

Now to calculate the molality

$$\dfrac{4.2880}{\mathrm{kg\ of\ solvent}} = \dfrac{4.2880}{\pu{1.07181 kg}} = 4.00\ \mathrm{molal}$$


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