This problem came up while solving some problems on mass percentage and mole concept.

Essentially, our teacher said that the $w/w\%$ and the $v/v\%$ for a given solution will be the same, because in order to obtain $v/v\%$ there will be a mutual division by density.

However, the density of solute is different from density of solution, right? For example, the problem we were solving involved $\ce{NaOH}$ dissolved in water. But $\ce{NaOH}$ as a compound has a characteristic density, different from the density of the solution. So $w/w$ and $v/v$ should have different values, right?

I'm not sure if I'm able to clarify what I'm thinking, but I need help with this nonetheless.


closed as unclear what you're asking by Mithoron, Mathew Mahindaratne, Jon Custer, Todd Minehardt, M.A.R. Aug 14 at 19:05

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  • 3
    $\begingroup$ Maybe your teacher meant %w/v, otherwise no, it doesn't make sense. $\endgroup$ – Buck Thorn Aug 8 at 18:32
  • $\begingroup$ Teacher said something wrong anyway $\endgroup$ – Alchimista Aug 9 at 7:44

The density of the solution isn't needed for either wt/wt% nor vol/vol%.

wt% = (grams solute)/(grams solution)

vol% = (vol solute)/(vol solution)

For vol% it only really makes sense if mixing two liquids like ethanol and water.


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