1
$\begingroup$

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.

Hence

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

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

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

Many thanks.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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