# Given a 1*L volume of solution that is 1.25*mol*L^-1 with respect to MgCl_2, there are...? "a. 2 moles of ions in solution." "b. 1.25 moles of ions in solution." "c. 2.5 moles of ions in solution." "d. 2 moles of ions in solution."

## Given a $1 \cdot L$ volume of solution that is $1.25 \cdot m o l \cdot {L}^{-} 1$ with respect to $M g C {l}_{2}$, there are... $\text{a. 2 moles of ions in solution.}$ $\text{b. 1.25 moles of ions in solution.}$ $\text{c. 2.5 moles of ions in solution.}$ $\text{d. 2 moles of ions in solution.}$

$\text{Option c............}$
The $\text{mole}$, ${N}_{A}$, is simply a number. Admittedly, it is a very large number, ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.
And if you have a mole of stuff, you have $6.022 \times {10}^{23}$ INDIVIDUAL ITEMS of that stuff. Here you have a concentration of $1.25 \cdot m o l \cdot {L}^{-} 1$ $M g C {l}_{2}$. In $1 \cdot L$ of solution there are $1.25 \cdot m o l$ $M {g}^{2 +}$ ions, BUT ALSO $2 \times 1.25 \cdot m o l$ $C {l}^{-}$ ions.....
And thus in total you have $2.5 \times {N}_{A}$ $C {l}^{-}$ per litre of this solution. This is a basic, underlying principle of chemistry, and if you can develop this idea of chemical equivalence, where MASS and CONCENTRATION corresponds to numbers of molecules and atoms, you will save yourself a lot of trouble,