In the laboratory a student combines 37.6mL of a
0.231 M magnesium
chloridesolution with 12.6 mL of a
0.669 M calcium chloride
solution.
What is the final concentration of chloride anion
?
Concentration of mixture = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
where C1 --> Concentration of 1 component
V1-->volume of 1 component
C2 --> Concentration of other component
V2-->volume of other component
n1 --> number of particle from 1 molecule of 1st component
= 2 as 1 molecule of MgCl2 has 2 Cl- ion
n2 --> number of particle from 1 molecule of 2nd component
= 2 as 1 molecule of CaCl2 has 2 Cl- ion
To calculate concentration of Cl-, use:
C = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
C = (2*0.231*37.6+2*0.669*12.6)/(37.6+12.6)
C = 0.6819 M
Answer: 0.682 M
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