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Question

50 mL of 0.05 M BaCl2 (aq) solution and 100 mL of 0.1 M of KCl (aq) solutions are mixed. Calculate the molarity of Cl ions in the resulting solution.

A
0.33 M
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B
0.10 M
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C
0.66 M
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D
1.33 M
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Solution

The correct option is B 0.10 M
We have,
50 mL of 0.05 M BaCl2 solution and 100 mL of 0.1 M KCl solution.
Millimoles of Cl ions from BaCl2(n1)=2×M1V1n1=2×0.05×50=5 millimole
Since, 1 mol of BaCl2 gives 2 mol of Cl ions.
Where, M1 and V1 are the molarity and volume in mL of the given BaCl2 solution.
Similarly,
Millimoles of Cl ions from KCl(n2) = 1×M2V2
n2=1×0.1×100=10 millimole
Since, 1 mol of KCl gives 1 mol of Cl ions.
Where, M2 and V2 are the molarity and volume in mL of the given KCl solution.
Total millimoles of Cl ions = 5+10 = 15 millimoles
Total volume of solution = 50 + 100
= 150 mL
Now, the molarity is given by:
M=moles of solutevolume of solution(mL)×1000
M(Cl)=15×103mol150(mL)×1000=0.1 M
Therefore, the molarity of Cl ion will be 0.10 M


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