A scale challenge for your brilliant minds.

[Whymsical]

Lets take that magical number 1 with 100 zeroes at the end known as google and see what we can put it into.

My challenge/ curiosity for the ones who love calculations is this:
Not counting for gravity, and using any temperature you choose, how large would a cube of pure lead have to be to hold a google of atoms.

Please show your work so we can all be accurately enthralled by the results.

 

[Doc Al]

We'll breathlessly await your trenchant analysis.

 

[Khashishi]

I think the term is googol

 

[Vagn]

Lets take that magical number 1 with 100 zeroes at the end known as google and see what we can put it into.

My challenge/ curiosity for the ones who love calculations is this:
Not counting for gravity, and using any temperature you choose, how large would a cube of pure lead have to be to hold a google of atoms.

Please show your work so we can all be accurately enthralled by the results.

It's not a particularly hard problem, if you know the mass density, ρ, and mass per mole, μ, of lead at a given temperature, along with avogadro's number N.

First you divide the mass density by the mass per mole, to get the molar density, the number of moles per unit volume. You then multiply by Avogadro's number to get the number of atoms per unit volume. You then divide a googol by the result and get the volume required for 10¹⁰⁰ lead atoms.
For STP I got 1.2X10⁷⁴ m³, corresponding to a cube with sides of length 4.93x10²⁴m, or slightly over 1/100th of the diameter the observable universe, or a sphere of radius 3.06x10²⁴m.

Of course the situation is unphysical, as you'd never get that many atoms together of a pure element.

 

[Coffee_]

10^99 * 207(relative mass of lead)*1,6*10^(-27)kg = mass of that block

ρ = m / V <=> V = m / ρ <=> V = (2,0716 * 10^75) kg / 11340kg/m^3 <=>

V= 1,82*10^71m^3