LProduct of growth Calculation, Magnesium bromide (Salze-73) This LProduct of growth of One of them Magnesium bromide solution of which: 6.9·10-4 mol3/L3. Calculate of which:, How a lot Grams The Commission Magnesium bromide in a Litre Solving.

  Aufgabe: Salze-73    

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This LProduct of growth of One of them Magnesium bromide solution of which: 6.9·10^-4 mol³/L³. Calculate of which:, How a lot Grams The Commission Magnesium bromide in a Litre Solving.

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Lösung:

Notes The Participants Ions are Mg²⁺ Responsibility. Br^–

1. 1·MgBr₂ ⇄ 1·Mg²⁺ + 2·Br^–
2. Es Applicable Therefore, 2·c(MgBr₂) = 2·c(Mg²⁺) = c(Br^–)
3. a) where applicable c(Br^–) = 2·c(Mg²⁺)
4.     L_p = c(Mg²⁺)·c²(Br^–) = c(Mg²⁺)·(2c(Mg²⁺))² = 4c³(Mg²⁺)
5.     c(Mg²⁺) = c(MgBr₂) = ∛(L_p/4) = 0.055 mol/L
6.     m = c·V·M = 0.055mol/l · 1.0 L · 184.113 g/mol =10.24 g/L
7.    
8. b) where applicable c(Mg²⁺) = c(MgBr₂) = ½·c(Br^–)
9.     L_p = ½·c(Br^–) ·c²(Br^–) = ½·c³(Br^–)
       Responsibility. c(Br^–) =∛(2·L_p)
10.     c(Br^–) = 2c(MgBr₂)
11. This means that c(MgBr₂) = ½·c(Br^–) = ½·∛(2·L_p)
        = ∛(1/8)·∛(2·L_p) = ∛(2·L_p/8) = ∛(L_p/4)
12.     Es is given The Commission The equal to Lthe island How In the case of a): c(MgBr₂) = ∛(L_p/4) = 0.055 mol/L
13.     The Remains would be So what do you mean? also identical and This means that m=10.24 g/L

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