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The molar mass of sulfuric acid is $M(\ce{H2SO4}) = \pu{98.1 g/mol}$.
We have $\pu{100 mL}$ of a $0.1$ molar solution of sulfuric acid. What volume $V$ in $\mathrm{mL}$ of water, $\ce{H2O}$, do we have to add in order to obtain a solution that has $\pu{4.9 g/L}$ sulfuric acid?

I've done the following:

\begin{align} M(\ce{H2SO4}) &= \pu{98.1 g/mol}\\ M(\ce{H2O}) &= \pu{18.0 g/mol}\\ \text{mass concentration of } H_2SO_4 &= \pu{98.1 g/L}\\ c(\ce{H2SO4}) &= \pu{1 mol/L}\\ c_{\mathrm{contr}} \times V_{\mathrm{contr}} &= c_{\mathrm{dil}}\times V_{\mathrm{dil}}\\ \Longrightarrow V_{\mathrm{dil}} &= \frac{c_{\mathrm{conctr}}\times V_{\mathrm{conctr}}}{c_{\mathrm{dil}}} \end{align}

The molarity of the diluted water I can't calculate with elementary formulas. So it leaves me with 2 unknowns.

I've tried using so many formulas to no avail.

What formula am I not thinking of? Or what variable am I missing?

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    $\begingroup$ Can the downvoter please explain his action? $\endgroup$ – Anonymous196 Oct 2 '18 at 14:44
  • $\begingroup$ 1) The body says "0.1 molar solution", while later you write c(H2SO4) = 1 mol/L; shouldn't the latter be 0.1 mol/L? 2) Where did you get the value for the density (rho) from? Did you just re-use the value of molar mass and change the the units? Don't do that :) You have to look up the density in tables, if it isn't given. You will then notice that the density changes in dependence of the concentration - so you had to note down, which H2SO4 concentration you are referring to when writing down the density. Here is an example tabular $\endgroup$ – Arsak Oct 3 '18 at 10:32
  • $\begingroup$ @Marzipanherz I'll change that. Yeah I shouldn't use the same notations as in physics. I'll change that too. $\endgroup$ – Anonymous196 Oct 3 '18 at 11:16
  • $\begingroup$ I see. Are you aware that this mass concentration is also referring to a 1mol/L solution, not to a 0.1molar solution the question talks about? $\endgroup$ – Arsak Oct 3 '18 at 12:05
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We have talked about it in chat and maybe the question is a bit ambiguous. I would interpret the last line as a mass concentration. So what you are actually looking to create is a $\gamma(\ce{H2SO4}) = \pu{4.9 g/l}$ solution. Then the calculation becomes straight forward.
Keep in mind that water is your solvent, so you can't actually dilute it; also writing down the concentration is a bit pointless here.

You can easily convert the mass concentration to an amount concentration. The relative molar mass $M$ is given by the mass $m$ of the substance divided by the amount of substance $n$: $$M = \frac{m}{n}\tag1\label{molarmass}$$

The mass concentration is defined as $$\gamma = \frac{m}{V}.\tag2\label{massconc}$$ The amount concentration is defined as $$c = \frac{n}{V}.\tag3\label{amountconc}$$ You can use \eqref{molarmass} in \eqref{amountconc} and obtain $$c = \frac{m}{M} \times \frac{1}{V} =\frac{m}{V} \times \frac{1}{M}.\tag4$$ Substitute \eqref{massconc} back in and you have $$c = \frac{\gamma}{M}.\tag5\label{amounttomass}$$

Now in the question you have given the following: \begin{align} M(\ce{H2SO4}) &= \pu{98.1 g/mol}\\ c_\mathrm{i}(\ce{H2SO4}) &= \pu{0.1 mol/l}\\ V_\mathrm{i}(\ce{H2SO4}) &= \pu{100 ml}\\ \gamma_\mathrm{f}(\ce{H2SO4}) &= \pu{4.9 g/l} \end{align} I have given the initial values the subscript $\mathrm{i}$ and the final values the subscript $\mathrm{f}$ (instead of concentrated and diluted).

You are now looking for the volume of solvent you have to add, which is $$\Delta V = V_\mathrm{f} - V_\mathrm{i}.\tag5\label{volume}$$

Then you are right, you need the following relation $$c_\mathrm{i} \times V_\mathrm{i} = c_\mathrm{f} \times V_\mathrm{f}, \tag6$$ rearranged and using \eqref{amounttomass} \begin{align} c_\mathrm{i} \times V_\mathrm{i} &= \frac{\gamma_\mathrm{f}}{M} \times V_\mathrm{f},\\ V_\mathrm{f} &= \frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i}, \end{align} and using \eqref{volume} \begin{align} V_\mathrm{i} + \Delta V &= \frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i},\\ \Delta V &= \frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i} - V_\mathrm{i}\\ &= \left(\frac{M \times c_\mathrm{i}}{\gamma_\mathrm{f}} - 1\right)V_\mathrm{i}. \tag7\label{dilute} \end{align}

Now simply plug in your values and get the solution (hover over box to reveal):

$\displaystyle \Delta V = \left(\frac{\pu{98.1 g//mol} \times \pu{0.1 mol//l}}{\pu{4.9 g//l}} - 1\right)\times \pu{0.100 l} = \pu{0.100 l}.$

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