
Assertion: In the metallurgy of aluminium, purified \[A{{l}_{2}}{{O}_{3}}\] is mixed with \[N{{a}_{3}}Al{{F}_{6}}\] or \[Ca{{F}_{2}}\].
Reason: \[N{{a}_{3}}Al{{F}_{6}}\] or \[Ca{{F}_{2}}\] lowers the melting point of mixture and increases its conductivity.
A) Both Assertion and reason are correct and Reason is the correct explanation of Assertion.
B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
C) Assertion is correct but Reason is incorrect.
D) Both Assertion and Reason are incorrect.
Answer
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Hint: In Hall-Heroult process, aluminium (\[Al\]) metal is extracted by the electrolytic reduction of alumina (\[A{{l}_{2}}{{O}_{3}}\]) by dissolving in molten cryolite or fluorspar to lowers the melting of the mixture and it also increases the conductivity of the electrolyte.
Complete answer:
> We know that in metallurgy aluminium is extracted through a process called Hall-Heroult. In this process aluminium is extracted by the electrolytic reduction of alumina in molten cryolite or fluorspar.
> Aluminium is very big to remove from its ore by the reduction of carbon. So, to remove aluminium there is a requirement of high temperature. It means cryolite and fluorspar lowers the melting point and increases the conductivity of the solution.
Now, coming to given options, option A, Both Assertion and reason are correct and Reason is the correct explanation of Assertion.
Therefore option A is correct because alumina is purified by mixing with \[N{{a}_{3}}Al{{F}_{6}}\] or \[Ca{{F}_{2}}\] and the reason is \[N{{a}_{3}}Al{{F}_{6}}\] or \[Ca{{F}_{2}}\] lowers the melting point of mixture and increases the conductivity of the electrolyte.
So, the correct option is A.
Note: Don’t be confused with the names cryolite and fluorspar. Both are not the same. The molecular formula of cryolite is \[N{{a}_{3}}Al{{F}_{6}}\] and the molecular formula of fluorspar is \[Ca{{F}_{2}}\] .
Both are ionic compounds and ionize in a suitable solvent gives ions and increases the electric conductivity of the solution.
Complete answer:
> We know that in metallurgy aluminium is extracted through a process called Hall-Heroult. In this process aluminium is extracted by the electrolytic reduction of alumina in molten cryolite or fluorspar.
> Aluminium is very big to remove from its ore by the reduction of carbon. So, to remove aluminium there is a requirement of high temperature. It means cryolite and fluorspar lowers the melting point and increases the conductivity of the solution.
Now, coming to given options, option A, Both Assertion and reason are correct and Reason is the correct explanation of Assertion.
Therefore option A is correct because alumina is purified by mixing with \[N{{a}_{3}}Al{{F}_{6}}\] or \[Ca{{F}_{2}}\] and the reason is \[N{{a}_{3}}Al{{F}_{6}}\] or \[Ca{{F}_{2}}\] lowers the melting point of mixture and increases the conductivity of the electrolyte.
So, the correct option is A.
Note: Don’t be confused with the names cryolite and fluorspar. Both are not the same. The molecular formula of cryolite is \[N{{a}_{3}}Al{{F}_{6}}\] and the molecular formula of fluorspar is \[Ca{{F}_{2}}\] .
Both are ionic compounds and ionize in a suitable solvent gives ions and increases the electric conductivity of the solution.
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