Respuesta :
Answer: The concentration of [tex]Cl^-[/tex] ions in the resulting solution is 1.16 M.
Explanation:
To calculate the molarity of the solution after mixing 2 solutions, we use the equation:
[tex]M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of the [tex]CaCl_2[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of the [tex]AlCl_3[/tex]
We are given:
[tex]n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL[/tex]
Putting all the values in above equation, we get
[tex]M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M[/tex]
The concentration of [tex]Cl^-[/tex] ions in the resulting solution will be same as the molarity of solution which is 1.16 M.
Hence, the concentration of [tex]Cl^-[/tex] ions in the resulting solution is 1.16 M.
The molarity of Cl¯ in the solution made by mixing 155 mL of 0.276 M CaCl₂ with 384 mL of 0.471 M AlCl₃ is 1.165 M
We'll begin by calculating the number of mole of chloride ion, Cl¯ in each solution.
For CaCl₂:
Volume = 155 mL = 155 / 1000 = 0.155 L
Molarity = 0.276 M
Mole of CaCl₂ =?
Mole = Molarity x Volume
Mole of CaCl₂ = 0.276 × 0.155
Mole of CaCl₂ = 0.04278 mole
CaCl₂(aq) —> Ca²⁺(aq) + 2Cl¯(aq)
From the balanced equation above,
1 mole of CaCl₂ contains 2 moles of Cl¯
Therefore,
0.04278 mole of CaCl₂ will contain = 0.04278 × 2 = 0.08556 mole of Cl¯
Thus, 0.08556 mole of Cl¯ is present in 155 mL of 0.276 M of CaCl₂.
For AlCl₃:
Volume = 384 mL = 384 / 1000 = 0.384 L
Molarity = 0.471 M
Mole of AlCl₃ =?
Mole = Molarity x Volume
Mole of AlCl₃ = 0.471 × 0.384
Mole of AlCl₃ = 0.180864 mole
AlCl₃(aq) —> Al³⁺(aq) + 3Cl¯(aq)
From the balanced equation above,
1 mole of AlCl₃ contains 3 moles of Cl¯
Therefore,
0.180864 mole of AlCl₃ will contain = 0.180864 × 3 = 0.542592 mole of Cl¯
Thus, 0.542592 mole of Cl¯ is present in 384 ml of 0.471 M of AlCl₃
- Next, we shall determine the total mole of Cl¯ in the resulting solution.
Mole of Cl¯ in CaCl₂ = 0.08556 mole
Mole of Cl¯ in AlCl₃ = 0.542592
Total mole = 0.08556 + 0.542592
Total mole = 0.628152 mole
- Next, we shall determine the total volume of the resulting solution
Volume of CaCl₂ = 0.155 L
Volume of AlCl₃ = 0.384 L
Total volume = 0.155 + 0.384
Total volume = 0.539 L
- Finally, we shall determine the molarity of Cl¯ in the resulting solution
Total mole = 0.628152 mole
Total volume = 0.539 L
Molarity of Cl¯ =?
Molarity = mole / Volume
Molarity of Cl¯ = 0.628152 / 0.539
Molarity of Cl¯ = 1.165 M
Therefore, the molarity of Cl¯ in the resulting solution is 1.165 M
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