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Question

# 50 mL of 0.05 M BaCl2 (aq) solution and 200 mL of 0.5 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
0.06 M
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C
0.60 M
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D
0.03 M
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Solution

## The correct option is B 0.06 MWe have, 50 mL of 0.05 M BaCl2 solution and 200 mL of 0.5 M KCl solution. Millimoles of Cl− ions from BaCl2, (n1)=2×M1V1∴n1=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=1×0.5×200=100 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+100=15 millimoles Total volume of solution =(50+200) mL=250 mL Now, the molarity 0f Cl− can be calculated: M(Cl−)=15×10−3 mol250 (mL)×1000=0.06 M Therefore, the molarity of Cl− ion will be 0.06 M

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