Mass of an atomic layer of a baloon

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To estimate the mass of aluminum required for a single atomic layer on a toy balloon, the balloon is modeled as a sphere with a radius of 10 cm. The surface area is calculated, and the volume of the atomic layer is determined by multiplying the surface area by the approximate radius of an aluminum atom. The mass is then found by using the volume and the density of aluminum, which is 2.7 x 10^3 kg/m^3. The discussion highlights a potential oversight in not incorporating the molar mass of aluminum, which is 27 grams per mole, into the calculations. The method proposed effectively uses density and volume to find mass, but further clarification on the role of molar mass is sought.
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An inflated smooth toy balloon is coated with a single atomic layer of aluminium. Esitimate the mass of Al that would be required. The density of Al is 2.7 x10^3kg/m^3 and its molar mass if 27 grams for mol.


I assusmed the balloon was a sphere of radius 10cm, found surface area then mutliplied by approximate radius of an atom to find volume of the layer. Then used volume x density to find mass, but this method doesn't make use of the molar mass in the question? Any other methods you can see?
 
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Myrddin said:
I assumed the balloon was a sphere of radius 10cm, found surface area then multliplied by approximate radius of an atom to find volume of the layer. Then used volume x density to find mass, but this method doesn't make use of the molar mass in the question? Any other methods you can see?

The approximate radius of the aluminium atom follows from the molar mass and density.

ehild
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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