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Question

100 mL of a 0.1 MNaCl solution is added to 300 mL of a 0.2 M MgCl2 solution. 100 mL of this solution was further diluted with 400 mL of water. The molarity of Cl ions in the resultant solution is:

A
0.725 M
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B
0.055 M
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C
0.045 M
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D
0.065 M
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Solution

The correct option is D 0.065 M
Millimoles of NaCl=MNaCl×VNaCl=100×0.1=10
Millimoles of Cl=10
Millimoles of MgCl2=MMgCl2×VMgCl2=300 mL×0.2 M=60
1 millimole of MgCl2 contains 2 millimoles of Cl
∴ Millimoles of Cl in the given MgCl2 solution =2×60=120
Total number of millimoles of Cl=10+120=130
Total volume after mixing =(100+300) mL=400 mL
Molarity of Cl on mixing these solutions =130400 M
Now, 100 mL of this solution is diluted with 400 mL water,
130400×100=Mf×500, where Mf is the final molarity of the resultant solution.
On solving, Mf=0.065 M

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