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Question

# 100 mL of a 0.1 M NaCl 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 MMillimoles 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 molarity of the resultant solution. Mf = 0.065 M

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