1)
[tex]C(ZnCl _{2} )= \frac{n(ZnCl _{2} )}{V} \\
C(ZnCl _{2} )= \frac{ \frac{m}{M(ZnCl _{2} )} }{V} \\
C(ZnCl _{2} )= \frac{ \frac{m}{M(Zn )+2M(Cl _{2})} }{V} \\
C(ZnCl _{2} )= \frac{ \frac{6,8g}{[65,4+2.(35,4)]g/mol} }{125.10 ^{-3}l } \\
C(ZnCl _{2} )= 0,4 mol/l[/tex]
2)
ZnCl₂ ---(H₂O)→ Zn²⁺ + 2Cl⁻
3)
n(Zn²⁺) = n(ZnCl₂) = 6,8/(65,4+2×35,4) = 0,04 mol
n(Cl⁻) = 2n(ZnCl₂) = 2×0,04 =0,08 mol
4)
C(Zn²⁺) = C(ZnCl₂) = 0,4 g/mol
C(Cl⁻) = 2n(ZnCl₂) = 2×0,4 =0,8g/mol