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Fittoniya [83]
3 years ago
10

What is the molarity (M) of chloride ions in a solution prepared by mixing 155 ml of 0.276 M calcium chloride with 384 ml of 0.4

71 M aluminum chloride?
Chemistry
1 answer:
sesenic [268]3 years ago
6 0

Answer: The concentration of Cl^- 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:

M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of the CaCl_2

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of the AlCl_3

We are given:

n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL  

Putting all the values in above equation, we get

M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M

The concentration of Cl^- ions in the resulting solution will be same as the molarity of solution which is 1.16 M.

Hence, the concentration of Cl^- ions in the resulting solution is 1.16 M.

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