Respuesta :

Wrys
(.11M NaCl)*(1 mol Cl/ 1 mol NaCl) = .11M from NaCl
(.11M MgCl2)*(2 mol Cl/ 1 mol MgCl2) = .055M from MgCl2

.11M + .055M = 0.165M of Cl

The concentration of Chloride ions in the solution of NaCl and Magnesium chloride has been 0.165 M.

The dissociation of the compound in the solution has been resulted in the formation of the constituent ions.

The concentration of the ions in the sample has been determined by the stoichiometric coefficient of the balanced equation.

Computation for concentration of Chloride ions

  • The concentration of Cl ions from sodium chloride solution has been given by:

[tex]\rm NaCl\;\rightarrow\;Na^+\;+\;Cl^-[/tex]

The concentration of Cl ions has been:

[tex]\rm 1\;M\;NaCl=1\;M\;Cl^-\\ 0.11\;M\;NaCl= 0.11\;M\;Cl^-[/tex]

The Cl ions from NaCl has been  0.11 M.

  • The Cl ions from magnesium chloride have been given as:

[tex]\rm MgCl_2\;\rightarrow\;Mg^2^+\;+\;2\;Cl^-[/tex]

The concentration of Cl ions has been:

[tex]\rm 1\;M\;NaCl=0.5\;M\;Cl^-\\ 0.11\;M\;NaCl= 0.11\;\times\;0.5\;M\;Cl^-\\0.11\;M\;NaCl=0.055\;M\;Cl^-[/tex]

The Cl ions from Magnesium chloride have been  0.055 M.

  • The total concentration of Cl ion has been:

[tex]\rm Cl^-=NaCl\;+\;MgCl_2\\Cl^-=0.11\;+\;0.055\;M\\Cl^-=0.165\;M[/tex]

The concentration of Chloride ions in the solution of NaCl and Magnesium chloride has been 0.165 M.

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