# Question #46250

Nov 26, 2016

Would you recheck your question? Given these conditions, $\text{concentration} = 14.4 \cdot m e q \cdot {L}^{-} 1$

#### Explanation:

First, we calculate $\left[M {g}^{2 +}\right]$ in units of $m o l \cdot {L}^{-} 1$.

$\left[M {g}^{2 +}\right] = \frac{0.675 \cdot g}{24.305 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{1925 \times {10}^{-} 3 \cdot L}$

$= 1.442 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

$1.442 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1 \times 1000 \cdot m e q \cdot m o {l}^{-} 1$

$= 14.42 \cdot m e q \cdot {L}^{-} 1$

$\left(i . e . \text{1 mole} = 1000 \times {10}^{-} 3\right)$

I am really not very happy in using this mass to represent a quantity of $M {g}^{2 +}$. It should have been a mass of a neutral salt, i.e. $M g C {l}_{2}$ or $M g {\left(N {O}_{3}\right)}_{2}$, which speciates in solution to $M {g}^{2 +}$ etc. I would check your question.