You need to prepare $\pu{250.0 mL}$ of an aqueous solution of ethanol in the concentration will be $\pu{0.500 mol L-1}$. You have $\pu{100 mL}$ of a commercial solution of ethanol in water $(35.0\% \,V/V)$ available. The density of ethanol (pure) under these conditions is $\pu{0.7893 g mL-1}$ and that of water is $\pu{0.9970 g / mL}$.

1.What volume (in $\mathrm{mL}$) will you need to withdraw to prepare this solution?

2.What is the mole fraction of ethanol in the resulting solution?

Answer to (1)

We can prepare $\pu{250 mL}$ of $\pu{0.5 mol L-1}$ solution of ethanol by diluting $\pu{20.85 mL}$ of $35 \%$ $\ce{C2H5OH}$ (density $\pu{0.7893 g/mL}$) with distilled water to a volume of $\pu{250 mL}$.

The dilution is therefore made by adding $\pu{229.15 mL}$ of water.

However, the volume of a liquid obtained by mixing measured volumes of different solutions is not always precisely the sum of the component volumes.


Source (B): $35 \%$ solution of $\ce{C2H5OH}$ (density $\pu{0.7893 g/mL}$)

The molar concentration of a solution of a chemical species A is the number of moles of that species contained in one litre of the solution (not in one litre of the solvent).

$$C(A) = \pu{0.5 mol/L}$$

$$\text{Molar concentration} = \frac{\text{Density} \cdot \text{Weight percent} }{ 100\cdot \text{Molecular weight}}$$

$$C(B) = \frac{d(B) \cdot w(B)}{100 \cdot M(B)}$$

$$C(B) = \frac{\pu{789.3 g/L} \times 35 \%}{100 \times \pu{46.0684 g/mol}}$$

$$C(B) = \pu{5.99662675499909 mol/L}$$

The number of moles of solute in the diluted solution must equal the number of moles in the concentrated reagent.

\begin{align} V(A) \cdot C(A) &= V(B) \cdot C(B) \\ V(B) &= \frac{V(A) \cdot C(A)}{C(B)} \\ V(B) &= \frac{\pu{0.25 L} \times \pu{0.5 mol/L}}{\pu{5.99662675499909 mol/L}} \\ V(B) &= \pu{2.08450525782339 \times 10^{-2} L} \\ V(B) &= \pu{20.85 mL} \\ \end{align}

Answer to (2)

We have

\begin{align} C(B) &= \frac{\pu{0.125029667842 mol}}{\pu{20.85 mL}} \\ &=\pu{5.99662675499909 mol/L} \end{align}

Now we have to add $\pu{229.15 mL}$ of water to $\pu{20.85 mL}$ of commercial solution of ethanol to make it $\pu{250 mL}$.

$\pu{229.15 mL}$ of water has a mass of $\pu{229.15 g}$. This means that there are $\pu{12.7197 mol}$ water because the molar mass of water is $\pu{18.0153 g}$

\begin{align} \text{Mole fraction of ethanol}&= \frac{\text{Number of moles of ethanol}}{\text{Number of moles of solution}} \\ &=\frac{\pu{0.125029667842 mol}}{(\pu{0.125029667842 mol} +\pu{12.7197 mol}}\\ & = \pu{0.009734 mol} \end{align}

Are my answers to both questions correct? If they are incorrect, where am I going wrong?

  • 3
    $\begingroup$ I guess if the time spent by writing this question were rather spent on double checking your procedure, you would know the answer even before posting it, not speaking about waiting for response. Note that the purpose of Chemistry SE is not proof reading, but understanding of principles, that cannot be easily get from textbooks/online resources. Is there any principle you have difficulties to apply to solve this particular task ? $\endgroup$
    – Poutnik
    Sep 25 '20 at 11:10
  • 2
    $\begingroup$ BTW, you should definitely consider reformatting the question using MathJax. $\endgroup$
    – Poutnik
    Sep 25 '20 at 11:11
  • 1
    $\begingroup$ What is the meaning of the 35.0% ethanol V/V ? The volumes are not additive. 35 mL pure ethanol and 65 mL pure water do not give 100 mL, but much less, maybe 96 mL. So is it 35 mL ethanol plus enough water to make 100 mL ? Also, it is ridiculous to make such calculations with 14 significant figures, if the result is required with 3 significant figures. $\endgroup$
    – Maurice
    Sep 25 '20 at 11:55
  • 1
    $\begingroup$ @Maurice 35%(V/V) concentration is defined as 35 mL of pure substance in the 100 mL of the final solution ( i.e. not 100 mL = 35 mL of ethanol + 65 mL of water - that would be volume fraction ) See en.wikipedia.org/wiki/Volume_fraction and en.wikipedia.org/wiki/Alcohol_by_volume $\endgroup$
    – Poutnik
    Sep 25 '20 at 12:18
  • $\begingroup$ @ Poutnik. Thank you ! $\endgroup$
    – Maurice
    Sep 25 '20 at 13:10

@ Dhamnekar. Your calculations are correct, provided your assumption (that the volumes are additive) is correct. Nevertheless note that the last value ($x = 0.0097$) has no unit : it is not in moles. And we must also admit that, due to your assumption, x is probably simply $x = 0.010$. Anyway, next time try to trust your calculations yourself. We are not here to verify correct calculations. We are here to HELP THOSE WHO DO NOT SUCCEED in solving their problems.

  • $\begingroup$ Thanks for pointing out that mole fraction don't have any unit. $\endgroup$ Sep 25 '20 at 13:42

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