# How do I calculate the following quantity? Volume of 2.106 M copper (II) nitrate that must be diluted with water to prepare 670.2 mL of a 0.8351 M solution?

Jun 30, 2017

$\text{Volume"="Moles"/"Concentration} = 266 \cdot m L$

#### Explanation:

We need $670.2 \cdot m L \times {10}^{-} 3 \cdot m L \cdot {L}^{-} 1 \times 0.8351 \cdot m o l \cdot {L}^{-} 1$

$= 0.660 \cdot m o l$ $\text{copper nitrate}$.

We have a $2.106 \cdot m o l \cdot {L}^{-} 1$ solution of $C u {\left(N {O}_{3}\right)}_{2} \left(a q\right)$ available....

And thus we take the quotient, $\text{volume} = \frac{0.660 \cdot m o l}{2.106 \cdot m o l \cdot {L}^{-} 1}$ $=$ $0.266 \cdot L = 265.7 \cdot m L$, and then dilute this volume up to $670.2 \cdot m L$.

You would usually never dilute solutions like this, so this question is not very practical and does not reflect standard lab practice. Ordinarily we would take $100 \cdot m L$ volumes or so of a mother solution, and dilute appropriately by a factor of 2 or 10.....

The important relationship was.....

$\text{Concentration"="Moles of solute"/"Volume of solution}$,

i.e. C("concentration")=(n("moles"))/(V("Litres")), and thus $n = C \times V$, $V = \frac{n}{C}$ etc.