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

Mar 23, 2017

$2.0 \cdot L$ of solution are required.

#### Explanation:

Look at the problem dimensionally:

we know that $\text{Moles of solute"/"Volume of solution} = 1.5 \cdot m o l \cdot {L}^{-} 1$

So $\text{Moles of solute"-="Volume of solution} \times 1.5 \cdot m o l \cdot {L}^{-} 1$

OR $\frac{\text{Volume of solution"="Moles of solute}}{1.5 \cdot m o l \cdot {L}^{-} 1}$

$= \frac{3.0 \cdot \cancel{m o l}}{1.5 \cdot \cancel{m o l} \cdot {L}^{-} 1} = 2 \cdot L$

We get an answer in $L$ because $\frac{1}{L} ^ - 1 = \frac{1}{\frac{1}{L}} = L$ as required.

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

$\text{Concentration}$ $=$ $\text{Moles of solute (mol)"/"Volume of solution (L)}$; and thus $\text{concentration}$ is commonly quoted with units of $m o l \cdot {L}^{-} 1$.

Capisce?

Mar 23, 2017

2 litres.

#### Explanation:

To actually work it out, use the simple relationship $M = \frac{n}{V}$

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

Then simply rearrange for V, so $V = \frac{n}{M}$ and plug in the numbers:

$V = \frac{n}{M} = \frac{3}{1.5} = 2$ litres