I was presented with the following problem in my lecture and I am confused as to what to do.

You are required to prepare $\pu{250 cm3}$ of $\pu{0.100 mol dm-3}$ HCl by diluting $\pu{0.600 mol dm-3}$ HCl with water.

Calculate the volume of $\pu{0.600 mol dm-3}$ HCl that must be diluted with water.

This was my attempt.

Moles of the $\pu{250 cm3}$ of $\pu{0.100 mol dm-3}$ HCl (target acid)

= $\frac{250}{1000} \times 0.100 = 0.0250$ moles.


$\frac{\pu{mol}}{\mathrm{conc}} = \mathrm{vol}_{\pu{cm3}} = \frac{0.0250}{0.600} \times 1000 = 41.66... \approx \pu{42.00 cm3}$

I'm not sure if the second step is right.

If I am right (hopefully) can someone please explain why.

Many thanks.


1 Answer 1


The clue is the moles in the dilute solution is the same as the moles of the concentrated stock solution - it has to be constant.

The method I have used is (therefore) correct.


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