We take 10mL of $H_2SO_4$ with a $96\%$ of mass and density $1.84g/cm^3$, and we add it into a flask that has $0.5L$ of capacity, filled up to half of the flask with distilled water. We shake the mixture and we add in more distilled water up to the level of $0.5L$. Find the molarity and molality.

$\mathbf{Attempt. }$ My problem comes up when trying to find molality, I've found molarity like this (and this meets with the book's solution):

Let $M$ be molarity, defined as $M=\frac{n_s}{V_{T}}$ where $n_s$ is the number of moles of the solute and $V_T$ the total volume of the mixture. As we know $V_s=0.01L$ and $d_s=1.84g/cm^3=1.84g/mL=1840g/L$. We can find the mass $m_s$: $$m_s=d_s\cdot V_s=18.4g$$ and then find the moles of the solute $n_s=18.4g\cdot\frac{1 mole\ H_2SO_4}{98g\ H_2SO_4}\approx 0.188 moles \ H_2SO_4$ and then the $V_T=0.5L$ so $$M=\frac{0.188moles}{0.5}\approx 0.38M$$ and this part agrees with my book now, the issue is with molality.

$\mathbf{Doubt/The\ part\ where\ I'm\ stuck.} $ With molality, which I'll denote $m$, defined as $m=\frac{n_s}{m_{solvent}(kg)}$, where $m_{solvent}$ is the mass of the solvent in kg. Okay well, I attempted this and it gave me $245.2m$ while the book's answer is $0.37m$. My attempt:

We know $n_s$, and $m_{solvent}=m_{mixture}-m_{solute}$, and from the percentage given of the solute $96\%$, we also know that $0.96m_{mixture}=m_{solute}\implies m_{mixture}=\frac{m_{solute}}{0.96}$, and we finally plug that in the previous equation: $m_{solvent}=\frac{m_{solute}}{0.96}-m_{solute}$, plugging $m_s=18.4g$ we get $m_{solvent}=0.767g$. Converting to kg, we get $m_{solvent}=0.000767kg$. And finally plugging into the molality: $$m=\frac{0.188moles}{0.000767kg}\approx 245.11m$$ where I have gone wrong so that it does not meet with my book's solution?

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    $\begingroup$ Your molarity calculation is somewhat incorrect. You forgot to take % wt/wt =96 into account. Your number are matching the book answer because of 96% but a similar calculation will fail for HCl which 37%. $\endgroup$
    – M. Farooq
    Jan 24 at 18:52
  • $\begingroup$ oh! okay but then how should I take it into account? Is it because the volume $10mL$ and density $1.84g/cm^3$ is actually of the whole mixture? Because if so, we can find the mass of the mixture, and take into account the 96% to find the actual mass of the solute $\endgroup$
    – Xetrez
    Jan 24 at 19:00
  • $\begingroup$ You know the density and you know that mass. But that mass is only 96% sulfuric acid. Multiply the calculated mass by 96/100. $\endgroup$
    – M. Farooq
    Jan 24 at 19:39
  • $\begingroup$ okay but that doesn't change a lot the molality, does it? I mean by multiplying 0.96*18.4 gives something less than 18 and it somewhat changes the molarity but not the molality and I'm still stuck on what did I do wrong on the molality $\endgroup$
    – Xetrez
    Jan 24 at 20:00
  • $\begingroup$ Okay, I will have a look at it. $\endgroup$
    – M. Farooq
    Jan 24 at 20:40

All in all this isn't a well stated problem. The molarity can be solved, but the molality can't really be solved.

The wording "with a 96% of mass" is odd. I'm not sure if this is your interpretation of the problem or if that is what is literately written in the book. I'll interpret the phrase to mean that the sulfuric acid was 96% sulfuric acid by weight.

For both problems the grams of sulfuric acid are:

$$\mathrm{grams\ }\ce{H2SO4} = 10 \times 1.84 \times 0.96 = 17.66$$

$$\mathrm{moles\ }\ce{H2SO4} = \dfrac{10 \times 1.84 \times 0.96}{98.079} = 0.1801 $$


So the molarity, $M$, of the solution will be:

$$\mathrm{Molarity} = \dfrac{10 \times 1.84 \times 0.96}{98.079 \times 0.50} = 0.360 $$


The problem with the molality calculation is that volumes are not additive. I'll admit that for two or three significant figures the calculation sort of works.

So for the 500 ml of solution, you must assume that 10 ml is sulfuric acid, the other 490 ml of volume is water. Water is assumed to be 1.00 kg/liter, so to figure out how many moles of sulfuric acid per kilogram of solvent we'll need to multiply the 0.360 moles per liter of solution by a correction factor of 1/0.980 which gives a value of 0.368 for the molality.


As a second way to calculate the molality let's divide moles of $\ce{H2SO4}$ by the kilograms of water.

From above we know that the 500 ml of solution contained 0.1801 moles of $\ce{H2SO4}$.

Again we have to assume that the volumes are additive so that 10 ml of concentrated sulfuric acid and 490 ml of water yield 500 ml of solution. Given no temperature we can just assume that water is 1.00 g/ml, so 490 ml of water has a mass of 490 grams or 0.490 kilograms.

$$\mathrm{Molality} = \dfrac{0.1801}{0.490} = 0.368 $$

To perhaps beat a dead horse here let's back calculate the volumes for one liter of a similar solution. Steffen's Chemistry Pages give that a 4% solution of sulfuric acid by weight has a density of 1.0250 g/ml, 41.00 g sulfuric acid per liter and a molarity of 0.418 at 20°C.

To get 41.00 grams of sulfuric acid we'd need

$$\mathrm{ml\ conc\ sulfuric\ acid} = \dfrac{41.00}{1.0250\times 0.96} = \pu{41.7 ml}$$

But since water is 1 g/ml and the solution is 96% water

$$\mathrm{ml water} = 1000 \times 0.96 \times 1.0250 = \pu{984 ml}$$

$$41.7 + 984 = 1025.7$$

  • $\begingroup$ I'm still unsure where I've used that the volumes are additive on my attempt to find the molality. Could you please pin-point it out? $\endgroup$
    – Xetrez
    Jan 24 at 21:39
  • $\begingroup$ @Xetrez - You messed up the molality calculation so badly I don't have a clue as to what you were doing. $\endgroup$
    – MaxW
    Jan 24 at 21:53
  • 1
    $\begingroup$ that doesn't help me much, I'd love to know what concept is wrong $\endgroup$
    – Xetrez
    Jan 24 at 22:11
  • $\begingroup$ @Xetrez - in your calculation is $$\mathrm{m_{solvent} = m_{mixture} − m_{solute}}$$ supposed to mean that $$\mathrm{molality(solvent) = molality(mixture) - molality(solute)}$$ $\endgroup$
    – MaxW
    Jan 24 at 22:22
  • 1
    $\begingroup$ I will elaborate on the problem a little bit in more detail. Your molarity part should have been solved. The problem is your original textbook/assignment question is that it does not tell us the total mass of the solvent for the molality part. If they had told us how much mass of water was added instead of "total" volume", the problem would have been better stated. $\endgroup$
    – M. Farooq
    Jan 25 at 0:26

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