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?

  • $\begingroup$ 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
    Commented Oct 3, 2018 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. $\endgroup$ Commented Oct 3, 2018 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
    Commented Oct 3, 2018 at 12:05

1 Answer 1


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):

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.