\(n_{NaCl}=\dfrac{5,85}{23+35,5}=0,1\left(mol\right)\)
Ta có:
\(A_{Cl}=\dfrac{35\cdot a+37\cdot\left(100-a\right)}{100}=35,5\\ =>a=75\%\\ =>\%_{37_{Cl}}=25\%\)
\(n_{37_{Cl}}=25\%\cdot0,1=0,025\left(mol\right)\)
Số nguyên tử của \(37_{Cl}=0,025\cdot6,22\cdot10^{23}=1,505\cdot10^{22}\)
Đáp án B