Question

$4.86\times {10}^x$ iron atoms are present in a stainless steel ball bearing having a radius of $0.254$ cm?The stainless steel contains $85.6$% Fe, by mass and has density $7.75$ g/mL.
So, the value of x is......

Answer 1

Volume of ball $d=4/3\pi \ast r^3$
So, mass of iron $=d\ast V\ast 0.856$
So, number of iron atoms in ball $=d\ast V\ast 0.856\ast N_o/56$ = $4.86\times {10}^{21}$
Therefore, the value of x is $21$.