Question

What volume of a `1.5*mol*L^-1` solution is required to deliver a `3.0*mol` quantity of solute?

Answer 1

`2.0*L` of solution are required.

Explanation:

Look at the problem dimensionally:

we know that `"Moles of solute"/"Volume of solution"=1.5*mol*L^-1`

So `"Moles of solute"-="Volume of solution"xx1.5*mol*L^-1`

OR `"Volume of solution"="Moles of solute"/(1.5*mol*L^-1)`

`=(3.0*cancel(mol))/(1.5*cancel(mol)*L^-1)=2*L`

We get an answer in `L` because `1/L^-1=1/(1/L)=L` as required.

In order to use these quotients, you have to remember the basic definition:

`"Concentration"` `=` `"Moles of solute (mol)"/"Volume of solution (L)"`; and thus `"concentration"` is commonly quoted with units of `mol*L^-1`.

Capisce?

Answer 2

2 litres.

Explanation:

To actually work it out, use the simple relationship `M = n/V`

Where M is molarity (or concentration) in moles per `dm^3` (or moles per litre, which is numerically identical), n is the number of moles of solute, and V is the volume of solution in `dm^3` (or litres, which are numerically identical).

Then simply rearrange for V, so `V = n/M` and plug in the numbers:

`V = n/M = 3/1.5 = 2` litres