(i) The nature of the solution is acidic. It is a mixture of conc. HNO₃, and conc. HCI in the ratio of 1 : 3. It can dissolve gold as it is an oxidising agent.
(ii) Pure gold is 24 carat gold which is very soft.

Metals high up in reactivity series are extracted by electrolytic reduction. While those in the middle are extracted first by converting into oxide and then reducing by carbon.
Same method cannot be used because metals have more affinity for oxygen than carbon.
Molten sodium chloride is taken for electrolytic reduction. The metals are deposited at the cathode and chlorine is liberated at anode.
At cathode: Na⁺ + e^- \(\longrightarrow \)Na
At anode: 2Cl^- \(\longrightarrow \) C1₂ + 2e^-

Take some metals like Ca, Mg and Fe etc. Place separately small pieces of these metals in beakers with small amount of water in them. If the metals react with the cold water, then the reaction is vigorous. If metal does not
react with cold water, then treat it with hot water. Record the observations and arrange the metals in decreasing order of reactivity.

(a) The main ore of aluminium is bauxite ore. Its formula is A1₂0₃.The impurities sociated with this ore are iron oxide and sand. .
(b) The reaction involved in obtaining pure alumina from mineral bauxite having impurities of iron oxide and silica are:
     (i) \(\underset { Aluminiumoxide\\ (frombauxiteore) }{ { A1 }_{ 2 }{ O }_{ 3 } } +\underset { Sodiumhydroxide\\ \quad (hot) }{ 2NaOH(aq) } \rightarrow \underset { Sodiumaluminate\\ \quad (Soluble) }{ { 2NaA1O }_{ 2 }(aq) } +\underset { Water }{ { H }_{ 2 }O(1) } \)
    The iron oxide and silica (sand) present in the bauxite ore do not dissolve in sodium hydroxide solution. Therefore, they are separated by filtration.
     (ii) \(\underset { Sodiumaluminate }{ { NaA1O }_{ 2 }(aq) } +\underset { Water }{ { 2H }_{ 2 }O(1) } \rightarrow \underset { Aluminiumhydroxide\\ \quad (Whiteppt) }{ { A1(OH) }_{ 3 }(s) } +\underset { Sodiumhydroxide }{ NaOH(aq) } \)
     (iii) \(\underset { Aluminiumhydroxide }{ { 2A1(OH) }_{ 3 }(s) } \underrightarrow { heat } \underset { Aluminiumoxide\\ \quad (Pure) }{ { A1 }_{ 2 }{ O }_{ 3 }(s) } +\underset { Steam }{ { 3H }_{ 2 }O(g) } \)

(a) The eating up of metals by the action of air and moisture on their surface is called corrosion.
      (i) Metal which does not corrode on being kept in the atmosphere is gold.
     (ii) Metal which corrodes on being kept in the atmosphere is iron.
(b) Take three boiling tubes and label them as A, B, and C. Place clean iron nails, free from rust, in each tube. Add water to tube A in such a manner that some parts of the nails are exposed to air and seal it with the help of a rubber cork. Add boiled water to tube B in such a way that the nails are completely submerged. Then, pour some amount of kerosene oil in the tube and seal it with the help of a rubber cork. Add anhydrous calcium chloride to tube C and then seal it. Leave the tubes undisturbed for two weeks and observe the changes. 

(a) The reaction involved in obtaining pure alumina from the mineral bauxite having impurities of iron oxide and silica is:
        (i) \(\underset { Aluminiumoxide\\ (frombauxiteore) }{ { AI }_{ 2 }{ O }_{ 3 }(s) } +\underset { Sodiumhydroxide\\ \quad (hot) }{ 2NaOH(aq) } \rightarrow \underset { Sodiumaluminate\\ \quad (Soluble) }{ { 2NaAIO }_{ 2 }(aq) } +\underset { Water }{ { H }_{ 2 }O(I) } \)
    The iron oxide and silica (sand) present in the bauxite ore do not dissolve in sodium hydroxide solution. Therefore, they are separated by filtration.
       (ii) \(\underset { Sodiumaluminate }{ { NaAIO }_{ 2 }(aq) } +\underset { Water }{ { 2H }_{ 2 }O(I) } \rightarrow \underset { Aluminiumhydroxide\\ \quad (Whiteppt) }{ { AI(OH) }_{ 3 }(s) } +\underset { Sodiumhydroxide }{ NaOH(aq) } \) 
       (iii) \(\underset { Aluminiumhydroxide }{ { 2AI(OH) }_{ 3 }(s) } \underrightarrow { heat } \underset { Aluminiumoxide\\ \quad (Pure) }{ { AI }_{ 2 }{ O }_{ 3 }(s) } +\underset { Steam }{ { 3H }_{ 2 }O(g) } \)
(b)

Ore A gives CO₂ and ore B gives SO₂ . Therefore, ores are MCO₃ and MS.
A can be converted into metal by the following steps :
\({ MCO }_{ 3 }\underrightarrow { Calcination } MO+{ CO }_{ 2 }\\ MO+C\underrightarrow { Reduction } M+CO\)
B can be converted into metal by the following steps :
\(2MS+{ 3O }_{ 2 }\underrightarrow { Roasting } 2MO+{ 2SO }_{ 2 }\\ \quad \quad \quad MO+C\rightarrow M+CO\)

Since the element B belongs to atomic number 17, it must be chlorine.
The element which burns golden flame in air is sodium. Therefore, the element A must be sodium.
When element sodium and chlorine reacts, it forms sodium chloride, the compound C. The compound C on electrolysis gives sodium hydroxide and hydrogen gas.
Thus, A = Na; B = CI₂; C = NaCI; D = NaOH.
         Chemical Reactions involved:
              \(2Na+{ CI }_{ 2 }\rightarrow 2NaCI\)
\(2NaCI(aq)+{ 2H }_{ 2 }O(I)\rightarrow 2NaOH(aq)+{ CI }_{ 2 }(g)+{ H }_{ 2 }(g)\)

Since the metal X react with cold water, it must be high in the reactivity series i.e., sodium or potassium.
X is alkali metal, Na or K.
Since the metal Y reacts with hot water, it must be middle in the reactivity series i.e., magnesium or calcium.
Y is alkaline earth metal, Mg or Ca.
Since the metal X reacts with steam only, it must be low in the reactivity series i.e., iron Z is Fe.
Increasing reactivity series : Na > Mg > Fe.

Since the non-metal A is the largest costituent of air, it must be Nitrogen. When nitrogen is heated with hydrogen in the ratio of 1 : 3, it gives ammonia in presence of iron as a catalyst [Haber's process]. When nitrogen is heated with oxygen, it gives nitric oxide. When this oxide is passed through water in presence of air, it gives nitric acid which acts as a strong oxidising agent.
(a) A - N₂; B - NH₃; C - NO; D - HNO₃
(b) Element A belongs to Group - 15 of the Periodic Table.