# In a 1*mL volume of solution that is 0.10*mol*L^-1 with respect to CaCl_2(aq), what is the NUMBER of "ions in solution?"

Jun 11, 2017

There are $3 \times {10}^{-} 4 \cdot m o l \times {N}_{A}$ ions in solution.......

#### Explanation:

Now $\text{concentration}$ is given by the quotient:

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

And thus to get the $\text{moles of solute}$ we takes the product

$\text{Volume of solution "xx" concentration}$.......

And so here,

$\text{moles of calcium chloride}$ $=$

$1 \cdot m L \times {10}^{-} 3 L \cdot m {L}^{-} 1 \times 0.10 \cdot m o l \cdot {L}^{-} 1 = {10}^{-} 4 \cdot m o l$

WITH RESPECT TO THE SALT $C a C {l}_{2}$.

Why did we emphasize $\text{the salt}$? Because EACH formula unit of $C a C {l}_{2}$ contributes 3 ions to the solution, i.e.....

$C a C {l}_{2} \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} C {a}^{2 +} + 2 C {l}^{-}$

And so we have 3xx10^-4*molxx6.022xx10^23*mol^-1=?? ions??