What volume of $0.5\ \mathrm M$ HCl solution is needed to prepare a $500\ \mathrm{ml}$ $10\ \%$ concentration solution, whose density is $1.05\ \mathrm{g/cm^3}$.

I've started off with finding the mass of the final solution, which would be $500\ \mathrm{cm^3} \times 1.05\ \mathrm{g/cm^3} = 525\ \mathrm g$. Next I find the mass of the HCl in it, $52.5\ \mathrm g$. After that, the moles of HCl, approx. $1.4384\ \mathrm{mol}$.

But after that I'm not sure how to proceed. I've tried dividing the molarity by the moles to get the volume, but that didn't work. I think I might have to make a system of equations but I'm bad at math so if that's how it's supposed to be done, I'd like for someone to show me how.


As you have calculated, 500 ml of a 10 % HCl solution contain about 1.44 mol HCl. Thus the concentration is about 2.9 mol/l. Therefore, you cannot prepare this solution by diluting a solution with c = 0.5 mol/l HCl.

  • $\begingroup$ Thanks, I guess I've forgotten that sometimes there just aren't any solutions (pun not intended) for such things! $\endgroup$ – Augustas May 1 '19 at 8:10
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    $\begingroup$ There is a way, using the fact HCl forms high boiling azeotrop with water, near 20% of HCl. So one can by rectification column evaporate excessive water, cool it down and dilute to 500 ml. :-) $\endgroup$ – Poutnik May 1 '19 at 9:18

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