# If 541*g of calcium is reacted with water to give a 1*L volume of solution, what is [Ca^(2+)]?

$\text{Molarity"="Moles of solute"/"Volume of solution}$
And thus $\text{Molarity} = \frac{541 \cdot \cancel{g}}{40.00 \cdot \cancel{g} \cdot m o {l}^{-} 1} \times \frac{1}{1 \cdot L}$ $>$ $10 \cdot m o l \cdot {L}^{-} 1$.
Dimensionally you get an answer in $\frac{1}{L \cdot m o {l}^{-} 1} = m o l \cdot {L}^{-} 1$, because $\frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$, as required.