\(\left( N _0\right) B =\frac{320}{32}=10 \text { moles }\)
\(N _{ A }=\frac{\left( N _0\right)_{ A }}{(2)^{2 / 1}}=\frac{20}{4}=5\)
\(N _{ B }=\frac{\left( N _0\right)_{ B }}{(2)^{2 / 5}}=\frac{10}{2^4}=0.625\)
Total \(N =5.625\)
No. of atoms \(=5.625 \times 6.023 \times 10^{23}\)