Ferric Chloride solution weight

Question

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)?

Answer 1

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.

Answer 2

My scientist friend explained to me that a 40% solution means that in 1 liter of solution there are 400 grams of solute. In your case you would add 40 grams to some water, dissolve it, then top it up to 100ml total solution.

If you add the solid to the liquid you have don't know what the total volume will be - therefore you don't know what the %-age will be. That's why you make it up to the final specified volume.