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When I notice that ferric chloride is sold as $40\%$ solution w/v, does it mean weight / volume?

Is it a matter of for example adding $40$g of ferric chloride crystals to a water solution where the total volume will then be $100$ml?

Or do I have to workout via the molar mass of $\ce{FeCl3}$ to water? How do I calculate the weight of crystals necessary to create a $40\%$ solution of $\ce{FeCl3}$?

Thanks

EDIT:

The system seems to think this is a homework question (whatever that is). If I divide $100$ml of water (so $100$g) by $40\%$, I get $40$g of $\ce{FeCl3}$. If I dissolve $40$g of $\ce{FeCl3}$ in $100$ml of water, will I get a $40\%$ solution of $\ce{FeCl3}$? And a total volume of $114$ml? ($\ce{FeCl3}$ is $2.9\frac{\mathrm g}{\mathrm{cm^3}}$, so $40$g is approx $13.8$ml)?

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  • $\begingroup$ Welcome to Chemistry! This is a homework question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. $\endgroup$ – M.A.R. Jul 7 '16 at 14:12
  • $\begingroup$ What? That doesn't make any sense? $\endgroup$ – made2hack Jul 7 '16 at 14:59
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You're right in that 40% weight/volume means that you take 40 grams of ferric chloride and dissolve it in 100 mL of water. This is the definition of w/v(%) and there is no need to work out the molar concentration - the main benefit of this method of stating concentration.

The total volume will not be (volume of liquid + volume of solid) in most cases of dissolution. In most cases, you will observe very little change in the volume of the resulting solution as the solid molecules separate and fit into the spaces between the liquid molecules. You may even occasionally observe a decrease in volume as shown in this experiment.

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