Respuesta :
The molarity of the chloride ions in the final solution is 0.747 M.
Given:
1.60 g of solid NaCl to 50.0 mL of 0.100 M calcium chloride solution.
To find:
The molarity of chloride ions in the final solution.
Solution:
Molarity of the calcium chloride solution = 0.100 M
Volume of the calcium chloride solution = V = 50.0 mL
[tex]1mL=0.001 L\\\\V=50.0mL=50.0 \times 0.001 L\\\\= 0.050 L[/tex]
Moles of calcium chloride = n
[tex]0.100 M=\frac{n}{0.050 L}\\\\n=0.100 M\times 0.050 L=0.005 mol[/tex]
[tex]CaCl_2(aq)\rightarrow Ca^+(aq)+2Cl^-(aq)[/tex]
Moles of chloride ions in solution = [tex]2\times 0.005 mol=0.010 mol[/tex]
Mass of sodium chloride = 1.60 g
Moles of sodium chloride :
= [tex]\frac{1.60 g}{58.5 g/mol}=0.02735 mol[/tex]
[tex]NaCl(s)\rightarrow Na^+(aq)+Cl^-(aq)[/tex]
Moles of chloride ions coming from 1.60 grams of NaCl = 0.02735 mol
Total moles of chloride ion in the final solution:
[tex]= 0.010 mol+0.02735 mol=0.03735 mol[/tex]
The molarity of the chloride ion in the final solution:
[tex]=\frac{0.03735 mol}{0.050 L}=0.747M[/tex]
The molarity of the chloride ions in the final solution is 0.747 M.
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