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o-na [289]
3 years ago
11

Determine the concentrations of mgcl2, mg2 , and cl– in a solution prepared by dissolving 2.75 × 10–4 g mgcl2 in 1.75 l of water

. express all three concentrations in molarity. additionally, express the concentrations of the ionic species in parts per million (ppm).
Chemistry
2 answers:
inysia [295]3 years ago
8 0

Answer:


M of MgCl₂ = 1.65 × 10⁻⁶ M


M of Mg²⁺ = 1.65 × 10⁻⁶ M


M of Cl⁻ = 3.30 × 10⁻⁶ M



Explanation:



1) MgCl₂


Molarity = number of moles of solute / volume of solution in liters, M = n / V


n = mass in grams / molar mass


molar mass of MgCl₂ = 24.305 g/mol + 2(35.543 g/mol) = 95.211 g/mol


n = 2.75 × 10⁻⁴ g / 95.211 g/mol = 2.89×10⁻³ moles


⇒ M = n / V = 2.89×10⁻³ moles / 1.75 l = 1.65 × 10⁻⁶ M



2) Mg²⁺ and Cl⁻


Those are the ions in solution.


You assume 100% dissociation of the ionic compound (strong electrolyte).


Then the equation is: MgCl₂ → Mg²⁺ + 2Cl⁻


That means that 1 mol of MgCl₂ produces 1 mol of Mg²⁺ and 2 moles of Cl⁻.


That yields the same molarity concentration of Mg²⁺ , while the molarity concentration of Cl⁻ is the double.



So, the results are:


M of MgCl₂ = 1.65 × 10⁻⁶ M


M of Mg²⁺ = 1.65 × 10⁻⁶ M


M of Cl⁻ = 3.30 × 10⁻⁶ M

Anvisha [2.4K]3 years ago
7 0

Answer 1) : To calculate the concentration of number of moles of MgCl_{2}, by dissolving 2.75 X 10^{-4} of MgCl_{2} will be like this;


Number of moles of MgCl_{2} = (2.75 X 10^{-4} g of MgCl_{2}) X (1 mole of MgCl_{2} / 95.211 g of MgCl_{2})


= 2.88 X 10^{-6} moles of MgCl_{2} (g/L)


In ppm it will be 0.288 ppm


Answer 2) To find the moles of Mg ions in solution of MgCl_{2}, we need to find the moles of Mg present in the solution,


Here, moles of Mg = Moles of MgCl_{2}


So, 2.88 X 10^{-6} moles of MgCl_{2} = 2.88 X 10^{-6} moles of Mg. (g/L)


And in ppm it will be 0.288 ppm


Answer 3) For calculating the moles of Cl ions present in the solution of MgCl_{2}, we need to know how many moles are present in MgCl_{2},


We see 2 moles of Cl ions are present in solution of MgCl_{2}


Therefore, 2 moles of Cl = 1 mole of MgCl_{2}


So, We can multiply 2 with the molarity of MgCl_{2}, we get,


2 X (2.88 X 10^{-6})= 5.77 X 10^{-6} g/L


And in ppm it will be 0.577 ppm.

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