# When 0.50 liter of 12 M solution is diluted to 1.0 liter, what is the molarity of the new solution?

Jan 5, 2017

The new concentration is HALF that of the original.

#### Explanation:

$\text{Concentration (C)}$ $=$ $\text{moles of solute (n)"/"volume of solution (V)}$.

Since $C = \frac{n}{V}$, $n = C V$.

And thus ${n}_{\text{initial}}$ $=$ $0.50 \cdot \cancel{L} \times 12 \cdot m o l \cdot \cancel{{L}^{-} 1} = 6 \cdot m o l$

But ${V}_{\text{final}}$ $=$ $1.0 \cdot L$.

So $\text{concentration}$

$=$ (0.50*cancelLxx12*mol*cancel(L^-1))/(1*L)=??mol*L^-1

You use the relationships, $C = \frac{n}{V}$, $V = \frac{n}{C}$, and $n = C V$ continually in a laboratory.

Concentrated hydrochloric acid is supplied as a $10 \cdot m o l \cdot {L}^{-} 1$ solution in water. If I have a $2.5 \cdot L$ bottle of conc. acid, how many litres of $1.0 \cdot m o l \cdot {L}^{-} 1$ can I prepare?

IMPORTANT: WE WOULD ALWAYS ADD CONC ACID TO WATER AND NEVER THE REVERSE!!