Answer :
1) Concentration NaCl
1.1- List the known quantities.
3.40 g NaCl
Volume: 43.3 mL
1.2- Molarity of NaCl
The molar mass of NaCl is 58.4428 g/mol.
[tex]mol\text{ }NaCl=3.40\text{ }g\text{ }NaCl*\frac{1\text{ }mol\text{ }NaCl}{58.4428\text{ }g\text{ }NaCl}=0.0582\text{ }mol\text{ }NaCl[/tex]Convert mL to L
1 L = 1000 mL
[tex]L=43.3\text{ }mL*\frac{1\text{ }L}{1000\text{ }mL}=0.0433\text{ }L[/tex]1.3- Set the equation
[tex]M=\frac{moles\text{ }of\text{ }solute}{liters\text{ }of\text{ }solution}[/tex]1.4- Plug in the known quantities
[tex]M=\frac{0.0582\text{ }mol\text{ }NaCl}{0.0433\text{ }L}=1.34\text{ }M\text{ }NaCl[/tex]The molar concentration of NaCl is 1.34 M NaCl.
2) Concentration of Na
The ion-molecule ratio is 1 mol Na = 1 mol NaCl.
[tex]M_{Na^+}=1.34\text{ }NaCl\text{ }\frac{mol\text{ }NaCl}{L}*\frac{1\text{ }mol\text{ }Na}{1\text{ }mol\text{ }NaCl}=1.34\text{ }*\frac{mol\text{ }Na^+}{L}[/tex]The concentration of Na+ is 1.34 M Na+.
3) Concentration of CaCl2
The ion-molecule ratio is 1 mol Ca (2+): 1 mol CaCl2.
Molarity: 0.556 M CaCl2
Volume: 0.0433 L
[tex]M_{Ca^{2+}}=0.556\text{ }\frac{mol\text{ }CaCl_2}{L}*\frac{1\text{ }mol\text{ }Ca^{2+}}{1\text{ }mol\text{ }CaCl_2}=0.556*\frac{mol\text{ }Ca^{2+}}{L}[/tex]The concentration of Ca2+ is 0.556 M Ca2+.
4) Concentration of Cl-
4.1- Moles of Cl- from NaCl
The ion-molecule ratio is 1 mol Cl = 1 mol NaCl.
Volume: 0.0433 L
The molar mass of NaCl is 58.4428 g/mol.
[tex]mol\text{ }NaCl=3.40\text{ }g\text{ }NaCl*\frac{1\text{ }mol\text{ }NaCl}{58.4428\text{ }g\text{ }NaCl}=0.0582\text{ }mol\text{ }NaCl[/tex]We have 0.0582 mol NaCl.
[tex]mol\text{ Cl}^-=0.0582\text{ }mol\text{ }NaCl*\frac{1\text{ }mol\text{ }Cl^-}{1\text{ }mol\text{ }NaCl}=0.0582\text{ }mol\text{ }Cl^-[/tex]We have 0.0582 mol Na from NaCl.
4.2- Moles of Cl- from CaCl2.
Molarity: 0.556 M CaCl2
Volume: 0.0433 L
[tex]M_{Ca^{2+}}=0.556\text{ }\frac{mol\text{ }CaCl_2}{L}*\frac{0.0433\text{ }L}{}=0.0241mol\text{ }CaCl_2[/tex]Moles of Cl-
0.0241 mol CaCl2.
The ion-molecule ratio is 2 mol Cl (-): 1 mol CaCl2.
[tex]mol\text{ }Cl^-=0.0241\text{ }mol\text{ }CaCl_2*\frac{2\text{ }mol\text{ Cl}^-}{1\text{ }mol\text{ }CaCl_2}=0.0482\text{ }mol\text{ }Cl^-[/tex]We have 0.0482 mol Cl- from CaCl2.
4.3- Add up moles of Cl- from NaCl and moles of Cl- from CaCl2.
We have 0.0582 mol Na from NaCl.
We have 0.0482 mol Cl- from CaCl2.
0.0582 + 0.0482 = 0.1064 mol Cl-
4.4- Molarity of Cl-
Moles of Cl-: 0.1064 mol
Volume: 0.0433 L
[tex]M=\frac{0.1064\text{ }mol\text{ }Cl^-}{0.0433\text{ }L}=2.46\text{ }M[/tex]The concentration of Cl- is 2.46 M Cl-.
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