How much water would we add to a #5.0moldm^-3# solution of #HCl(aq)# to make a #0.20moldm^-3# solution?

Answer 1

Well, #480*cm^3#....

Normally you add acid to water. Why? Because if you spit in conc. acid it spits back; I kid you not. Here, while the starting acid was not so concentrated, best and safe practice demands reverse addition, #"acid to water"#, and this is something your teacher should emphasize.

The volumes should reasonably be additive. And thus you would add #20*cm^3# of acid to #480*cm^3# of water....

And of course we could work out the new molar concentration given accurate volumes...

#"Concentration"="Moles of solute"/"Volume of solution"#

#=(20*cancel(cm^3)xx10^-3*cancel(dm^3*cm^-3)xx5.0*mol*cancel(dm^-3))/(500*cancel(cm^3)xx10^-3*L*cancel(cm^-3))=0.20*mol*L^-1#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How much water would we add to a #5.0*mol*dm^-3# solution of #HCl(aq)# to make a #0.20*mol*dm^-3# solution?

How much water would we add to a #5.0moldm^-3# solution of #HCl(aq)# to make a #0.20moldm^-3# solution?