
Cryolite is:
A-$N{a_3}Al{F_6}$ and is used in the electrolysis of alumina for decreasing electrical conductivity.
B-$N{a_3}Al{F_6}$ and is used in the electrolysis of alumina for lowering the melting point of alumina only.
C-$N{a_3}Al{F_6}$ and is used in the electrolysis of alumina for lowering the melting point and increasing the conductivity of alumina.
D-$N{a_3}Al{F_6}$ and is used in the electrolysis refining of alumina.
Answer
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Hint: Cryolite is a mineral that was found in Greenland in the year 1987. It was initially used as an ore of aluminum, but later it is used as a solvent in electrolytic processing of alumina where it plays an important role.
Complete step by step solution:
> It is an uncommon mineral that was identified on the west coast of Greenland in the year 1987. The molecular formula of cryolite is $N{a_3}Al{F_6}$. The chemical name of cryolite is sodium hexafluoroaluminate. It is generally found in large granular form. It is usually a white colour mineral, but cryolites with other colours (brown, red etc.) are also possible.
- Now, cryolite has been used as an ore of aluminum for a very long time. It is now used as one of the most important components in the electrolytic processing of aluminum oxide (also known as alumina)
-In the Hall-Heroult process, the molten cryolite is utilized as a solvent for aluminum oxide ($A{l_2}{O_3}$).
-The main objective of the Hall-Heroult process is to refine the aluminum from alumina ($A{l_2}{O_3}$).
-After dissolving the aluminum oxide, the molten cryolite decreases the melting point of the aluminum oxide by a decent margin.
-Apart from that, it also increases the conductivity of the dissolved aluminum oxide.
-As a result of these two, the extraction process of aluminum from alumina becomes easier and more economical.
-Hence, cryolite plays an important role in this Hall-Heroult process.
So, we can sum up by stating the fact that cryolite is nothing but $N{a_3}Al{F_6}$ and it mainly decreases the melting point of the alumina and increases the conductivity at the same time.
Let us now look at the answer options available regarding cryolite:
-It is $N{a_3}Al{F_6}$ , but it is not primarily used for decreasing the electrical conductivity of alumina. Hence, option A cannot be true.
-It is $N{a_3}Al{F_6}$ , but it is not used in the electrolysis refining of alumina, rather it is used in the electrolytic processing of alumina from which aluminum can be refined. Hence, option D cannot be true.
-Between option B and option C, option B provides the partial correct answer (which is lowering the melting point of alumina), whereas, option C provides the complete answer (which is lowering the melting point and increasing the conductivity of alumina)
Hence, option C is the correct answer to this question.
Note: Students might get confused between refining alumina and refining aluminum that may lead to a wrong decision. The cryolite is used in an electrolysis process in which alumina is processed. Now, the aluminum is refined from alumina. Hence, in this electrolysis process, alumina is being refined, but aluminum is.
Complete step by step solution:
> It is an uncommon mineral that was identified on the west coast of Greenland in the year 1987. The molecular formula of cryolite is $N{a_3}Al{F_6}$. The chemical name of cryolite is sodium hexafluoroaluminate. It is generally found in large granular form. It is usually a white colour mineral, but cryolites with other colours (brown, red etc.) are also possible.
- Now, cryolite has been used as an ore of aluminum for a very long time. It is now used as one of the most important components in the electrolytic processing of aluminum oxide (also known as alumina)
-In the Hall-Heroult process, the molten cryolite is utilized as a solvent for aluminum oxide ($A{l_2}{O_3}$).
-The main objective of the Hall-Heroult process is to refine the aluminum from alumina ($A{l_2}{O_3}$).
-After dissolving the aluminum oxide, the molten cryolite decreases the melting point of the aluminum oxide by a decent margin.
-Apart from that, it also increases the conductivity of the dissolved aluminum oxide.
-As a result of these two, the extraction process of aluminum from alumina becomes easier and more economical.
-Hence, cryolite plays an important role in this Hall-Heroult process.
So, we can sum up by stating the fact that cryolite is nothing but $N{a_3}Al{F_6}$ and it mainly decreases the melting point of the alumina and increases the conductivity at the same time.
Let us now look at the answer options available regarding cryolite:
-It is $N{a_3}Al{F_6}$ , but it is not primarily used for decreasing the electrical conductivity of alumina. Hence, option A cannot be true.
-It is $N{a_3}Al{F_6}$ , but it is not used in the electrolysis refining of alumina, rather it is used in the electrolytic processing of alumina from which aluminum can be refined. Hence, option D cannot be true.
-Between option B and option C, option B provides the partial correct answer (which is lowering the melting point of alumina), whereas, option C provides the complete answer (which is lowering the melting point and increasing the conductivity of alumina)
Hence, option C is the correct answer to this question.
Note: Students might get confused between refining alumina and refining aluminum that may lead to a wrong decision. The cryolite is used in an electrolysis process in which alumina is processed. Now, the aluminum is refined from alumina. Hence, in this electrolysis process, alumina is being refined, but aluminum is.
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