Question

7.36 x 1024 free oxygen atoms

Answer 1

Solution.

$N=7.36\sdot 10^{24};$

$N_A=6.02\sdot 10^{23}mol^{-1};$

$\nu=\dfrac{N}{N_A};$

$\nu=\dfrac{7.36\sdot10^{24}}{6.02\sdot 10^{23}}=12.2 mol;$

Answer: $\nu=12.2 mol.$