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A solution of hydrochloric acid of mass fraction $\pu{18.0\%}$ has a density of $\pu{1.0878 g cm-3}$. What are the molar and mass concentrations of this solution?

I got $\pu{5.37e-3}$ and $\pu{195.8}$, but it looks wrong.

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closed as off-topic by Jan, Mithoron, ron, Todd Minehardt, paracetamol Nov 24 '17 at 17:40

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    $\begingroup$ No matter what you got, returning it without units is wrong. $\endgroup$ – Ivan Neretin Nov 24 '17 at 10:16
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    $\begingroup$ @andselisk You voted to keep this open, probably because the OP says he isn't sure if his answers(s) are right. But I'm under the impression that simply stating the final "answer(s)" beneath the question doesn't necessarily imply the OP put effort in it (perhaps the OP's comparing two options in a question). If the OP includes his working/calculations here, then that counts as effort IMO. Just thought I'd mention it (why I voted to close), no biggie O:). Crisp answer you've put up ;-) $\endgroup$ – paracetamol Nov 24 '17 at 17:49
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    $\begingroup$ @paracetamol I'm still under the impression of recent close statistics. I decided that OP actually put some effort to solve the problem after I realized the values are not copy-pasted from the answer in textbook or some internet forum and the approach to solve the problem was not entirely wrong. But thanks for investing your time in explanation and kind words:) $\endgroup$ – andselisk Nov 24 '17 at 22:49
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    $\begingroup$ @andselisk < Nods > I read ortho's post too... I lamented over it for a while, and then kinda forgot about it. But it's admirable that you're actually working towards the change that ortho (and the other Mods) envisioned, by being more tolerant (then again, you always were) to posts that most of us would close without a second thought. Kudos to you! < salutes > $\endgroup$ – paracetamol Nov 25 '17 at 5:58
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The values you've obtained signify that you are aware of the mathematical expressions for molar and mass concentrations, but your attitude towards handling units is somewhat frivolous. I would recommend to think about how these three ways of expressing concentrations are interconnected, and always perform calculations with respected units.

So, in order to find molar concentration $C_i$ and mass concentration $\gamma_i$ defined as

\begin{align} C_i &= \frac{n_i}{V} \\ \gamma_i &= \frac{m_i}{V} \end{align}

one needs to express the amount $n_i$ and mass $m_i$ via volume $V$, density $\rho$, mmass fraction $\omega_i$ and molecular weight $M_i$:

$$\omega_i = \frac{m_i}{m} \quad \to \quad m_i = \omega_i m$$ $$m = \rho V$$ $$m_i = \omega_i \rho V$$ $$n_i = \frac{m_i}{M_i} = \frac{\omega_i \rho V}{M_i}$$

Finally, the expressions are ready to plug the known values in:

$$\require{cancel} C_i = \frac{\omega_i \rho \cancel{V}}{M_i \cancel{V}} = \frac{\omega_i \rho}{M_i} = \frac{0.18 \cdot \pu{1.0878 g cm-3}}{\pu{36.48 g mol-1}} = \pu{5.37e-3 mol cm-3} = \pu{5.37 M}$$

$$\gamma_i = \frac{\omega_i \rho \cancel{V}}{\cancel{V}} = \omega_i \rho = 0.18 \cdot \pu{1.0878 g cm-3} = \pu{0.196 g cm-3} = \pu{196 g L-1}$$

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