Respuesta :
Answer:
0.3026 M
Explanation:
Whe we have start with the ionization reaction of [tex]CaCl_2[/tex], so:
[tex]CaCl_2_(_a_q_)~--->~2Ca^+^2~+~Cl^-^1[/tex]
The molar ratio between [tex]CaCl_2[/tex] and [tex]Cl^-^1[/tex] is 2:1. With this information we can calculate the concentration of chloride ions in the 0.1 M solution:
[tex]0.1~M~CaCl_2\frac{2}{1}=~0.2~M~Cl^-^1[/tex]
With the concentration of chloride ions and the volume we can calculate the total moles of [tex]Cl^-^1[/tex] (lets keep in mind that 50 mL = 0.05L):
[tex]0.2~M=\frac{#~mol}{0.05~L}[/tex]
[tex]mol~=~0.2*0.05=~0.01~mol~Cl^-^1[/tex]
We have so far 0.01 mol of [tex]Cl^-^1[/tex]. With the information of NaCl we can calculate the moles of [tex]Cl^-^1[/tex] present in 0.3 of NaCl (the molar mass of NaCl 58.44 g/mol):
[tex]0.3~g~NaCl\frac{1~mol~NaCl}{58.44~g~NaCl}=0.00513~mol[/tex]
Now we can calculate the total moles of [tex]Cl^-^1[/tex]:
[tex]0.00513~mol~+~0.01~mol=~0.01513~mol~Cl^-^1[/tex]
With the total moles and the volume (50mL= 0.05L) we can calculate the concentration:
[tex]M=\frac{0.01513~mol~Cl^-^1}{0.05~L}=~0.3026~M~of~Cl^-^1~ions[/tex]
The total concentration of [tex]Cl^-^1[/tex] ions is 0.3026 M
Answer:
The molarity of the chloride ion is 0.3026 M ≈ 0.303 M
Explanation:
Step 1: Data given
Mass of solid NaCl = 0.300 grams
Volume of CaCl2 = 50.0 mL
Molarity of CaCl2 solution = 0.100 M
Volume of the final solution = 50.0 mL
Step 2: The balanced equation
NaCl → Na+ + Cl-
CaCl2 → Ca^2+ + 2Cl-
CaCl2 → NaCl2 + CaCl
Step 3: Calculate the number of moles Cl2 in CaCl2
In 1 mol CaCl2 we have 1 mol Ca^2+ and 2 mol Cl-
Moles CaCl2 = molarity ¨volume
Moles CaCl2 = 0.100 M * 0.050 L
Moles CaCl2 = 0.005 moles
In 1 mol CaCl2 we have 1 mol Ca^2+ and 2 mol Cl-
For 0.005 moles CaCl2 we have 2*0.005 = 0.010 moles Cl-
Step 4: Calculate moles Cl in NaCl
Moles NaCl = mass NaCl / molar mass NaCl
Moles NaCl = 0.300 grams / 58.44 g/mol
Moles NaCl = 0.00513 moles
In 1 mol NaCl we have 1 mol Na+ and 1 mol Cl-
In 0.00513moles NaCl we have 0.00513 moles Cl-
Step 5: Calculate the total number of moles Cl-
Total moles Cl- = 0.010 moles + 0.00513 moles
Total moles Cl- = 0.01513 moles Cl-
Step 6: Calculate the molarity of Cl-
Molarity = moles / volume
Molarity Cl- = 0.01513 moles / 0.050 L
Molarity Cl- = 0.3026 M
The molarity of the chloride ion is 0.3026 M ≈ 0.303 M
