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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.

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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.

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  • $\begingroup$ Thanks, I guess I've forgotten that sometimes there just aren't any solutions (pun not intended) for such things! $\endgroup$ – Augustas May 1 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 at 9:18

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