Question #46250

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Question #46250

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Answer 1 · 2016-11-26T07:36:18

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

Explanation:

First, we calculate $[M g^{2 +}]$ $[Mg^(2+)]$ in units of $m o l \cdot L^{- 1}$ $mol*L^-1$ .

$[M g^{2 +}] = \frac{0.675 \cdot g}{24.305 \cdot g \cdot m o l^{- 1}} \times \frac{1}{1925 \times 10^{- 3} \cdot L}$ $[Mg^(2+)]=(0.675*g)/(24.305*g*mol^-1)xx1/(1925xx10^-3*L)$

$= 1.442 \times 10^{- 2} \cdot m o l \cdot L^{- 1}$ $=1.442xx10^-2*mol*L^-1$

$1.442 \times 10^{- 2} \cdot m o l \cdot L^{- 1} \times 1000 \cdot m e q \cdot m o l^{- 1}$ $1.442xx10^-2*mol*L^-1xx1000*meq*mol^-1$

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

$(i . e . 1 mole = 1000 \times 10^{- 3})$ $(i.e. "1 mole"=1000 xx10^-3)$

I am really not very happy in using this mass to represent a quantity of $M g^{2 +}$ $Mg^(2+)$ . It should have been a mass of a neutral salt, i.e. $M g C l_{2}$ $MgCl_2$ or $M {g (N O_{3})}_{2}$ $Mg(NO_3)_2$ , which speciates in solution to $M g^{2 +}$ $Mg^(2+)$ etc. I would check your question.

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Question #46250

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