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A drop (0.05ml) of 12M HCl is spread over a thin sheet of aluminum foil (thickness 0.10mm and density of Al=2.70g/ml).Assuming whole of the HCl is used to dissolve Al,what will be the maximum area of hole produced in foil?


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Meq of Al=Meq of HCl
$\qquad\quad\;\;=12\times 0.05=0.6$
Weight of Al=$\large\frac{0.6\times 9}{1000}$$=0.0054g$
Volume of Al foil=$\large\frac{0.0054}{2.7}$$cm^3$
Area$\times$ thickness=volume
Hence (b) is the correct answer.
answered Jan 7, 2014 by sreemathi.v

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