Firstly,  Ag\(^+\)  + e\(^-\) →  Ag

  1 mole = 1 F = 96,500C x 1 e\(^-\)

⇒ If  96,500C liberates  108g of Ag

  x will  liberate  16g of Ag

 x = \(\frac{96,500 x 16}{108}\)  =  14,296.30 C

i.e The quantity of electricity that would liberate 16g of Ag which is also the same quantity that will liberate X of Mg = 14,296.30 C

Thus, Mg\(^{2+}\)  + 2e\(^-\) →  Mg

 If 2e\(^-\)  X  96,500C  liberates 24g of Mg

  14,296.30C  will  liberate  Xg of Mg

By simple proportion, X = \(\frac{14,296.30 x 24}{193,000}\) 

  = 1.78g