Helium atom's Atomic Weight = He = 4.0026 g/mol.
Avagrado's number is = 6.023 x 10^23 molecules / mol.
In order to find the number of atoms in 535kg of helium blimp.
First we need to convert the weight of 535 kg to mol which can be done by multiplying the atomic weight of helium into present atom with respect to grams.
535 kg converted into grams ---> 535000 g
There fore 535000 g X 4.0026 g/mol = 133,663 mol. (Note The grams will get cancel as per multiplication rules)
Multiplying the avagadro's number with the equation we get:
133,663 mol *6.0221415 Ă— 10^23 molecules/mol= 8.04919303 Ă— 10^28 molecules.
Since Helium is having 2 Atoms :
8.04919303 Ă— 10^28 molecules *2= 1.60983861 Ă— 10^29 atoms
The No of Helium atoms in 535 kg of helium is 1.61 x 10^29.
