What is the number of copper atoms in 1naira coin which weighs 7.39g assuming the material of which the coil is made is 86% copper na=n/n number of moles/n

Answer is: 6,023·10²² atoms of copper.
m(coin) = 7,39g.
ω (Cu) = 86% ÷ 100% = 0,86.
ω - mass percentage of copper in coin.
m(Cu) = ω· m(coin) = 0,86·7,39g = 6,355g.
n(Cu) = m÷M =6,355g ÷ 63,55g/mol = 0,1mol.
N(Cu) = n·Na = 0,1·6,023 × 10²³ = 6,023·10²².
Na -  Avogadro's number.

Ithink its solid. hope this