\(n_{halogen}=\dfrac{35,875-18,625}{108-39}=0,25\left(mol\right)\\ M_{muối\left(t.gia\right)}=\dfrac{18,625}{0,25}=74,5=39+M_{halogen}\\ \Leftrightarrow M_{halogen}=35,5\left(\dfrac{g}{mol}\right)\\ Vậy.muối.halogen.đó:KCl\left(Cl=35,5\right)\\ Chọn.A\)