Published by:
CGP EDU Academic Team
Published on: August 14, 2026
A drop (0.05 mL) of 12 M HCl is spread over a thin sheet of aluminium foild (thickness 1 mm and density of Al = 2.7 g/mL). Assuming whole of the HCl is used to dissolve. At what will be the maximum area of hole produced in foild (in cm 2 ).
Text Solution
Verified by ExpertsThe correct answer is:
2
(2)
m moles of HCl = 12 × 0.05 = 0.6
Now Al + 3HCl → AlCl 3 + 
H 2
so m moles of Al = 
× 0.6
or weight of Al = 
× 
= 0.0054 gram
∴ Volume of foil = 
mL or cm 3 = 0.002 cm 3
Now , Area × thickness = Volume
∴ Area = 
= 0.2 cm 2 (thickness = 0.01 cm)
= 0.2 × 10 = 2
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