$ 6.5 $ gram Mg deposits on electrolysis of $ MgC{{l}_{2}} $ fused then what will be the volume of chlorine gas at STP discharged at electrode:
(A) $ 3.033\text{ }L $
(B) $ 6.067\text{ }L $
(C) $ 9.1\text{ }L $
(D) $ 12.133\text{ }L $

Hint: We know that electrolysis is a very important process in metallurgy. This process is often being used to get ultra-pure metals. For purification of different metals different processes are available. Thus, electrolysis of aqueous solution of carnallite does not give magnesium.

Complete answer:
As we know that for Magnesium, we have hall heroult process and serpeck process and for some highly reactive metals like sodium we directly get them by electrolysis of their ore like rock salt. When a compound which is made up of two or more elements has a certain fixed ratio of its elements no matter from which it comes, this is known as the law of constant proportion. So, we have to prove that the concentration of chlorine is the same from both sources.
The elements having reduction potential greater than the reduction potential of hydrogen give the same results for electrolysis in fused and aqueous states. The elements having reduction potential greater than the reduction potential of hydrogen give the same results for electrolysis in fused and aqueous states. The reduction potential for magnesium is less than the reduction potential of hydrogen. Thus, one mole of $ MgC{{l}_{2}} $ gives one mole of Mg of $ 24 $ g and one mole of $ C{{l}_{2}}. $
 $ \therefore 24 $ gram of $ Mg=22.4d{{m}^{3}} $ of $ C{{l}_{2}} $ gas at STP.
Thus, $ 6.5 $ g of $ Mg=\dfrac{22.4\times 6.5}{24}=6.067 $ L of $ C{{l}_{2}} $ gas.

Note:
Remember that the elements deposited on both the electrodes are always equal and this is actually Faraday's second law of electrolysis. So, if we have the information about the quantity of one element deposited, we can actually calculate the other element's quantity deposited or discharged on the other electrode.