How many ions result from the dissolution of a 11.1*g mass of calcium chloride in water?

Mar 27, 2017

Approx. $1.81 \times {10}^{23}$ ions

Explanation:

$\text{Moles of calcium chloride}$ $=$ $\frac{11.1 \cdot g}{110.98 \cdot g \cdot m o {l}^{-} 1} = 0.100 \cdot m o l$

And thus in solution:

$C a C {l}_{2} \left(s\right) + \text{excess } {H}_{2} O \rightarrow C {a}^{2 +} + 2 C {l}^{-}$, there are $0.100 \cdot m o l$ $C {a}^{2 +}$ ions, and $0.200 \cdot m o l$ $C {l}^{-}$ ions, i.e. $0.300 \cdot m o l$ of ions...........

And $0.300 \cdot m o l \times {N}_{A} = 0.300 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

$1.81 \times {10}^{23}$ ions..........