Question

State the function of fluorspar in the metallurgy of aluminium.
A.) to increase conductivity
B.) to decrease conductivity
C.) to increase ductility
D.) none of these

Answer 1

Hint: The fluorspar lowers the melting point of alumina ($A{l}_2{O}_3$) and as we know that the pure alumina($A{l}_2{O}_3$) is a poor conductor of electricity so addition of fluorspar increases the mobility of the solution.

Complete step by step answer:
As we know that aluminum is the most abundant metal on Earth and it occurs in nature in the form of aluminium oxide ($A{l}_2{O}_3$) and other combined forms. This aluminium oxide is first obtained from bauxite ore by Bayer's process. The metallurgy is the process of extraction of metals from their natural states. The metallurgy of aluminium is done by the process which is named as Hall-Heroult process. Hall- Heroult is the process of reduction of aluminum from aluminium oxide($A{l}_2{O}_3$) by the electrolytic process. As we know that the aluminium oxide ($A{l}_2{O}_3$) is a very poor conductor of electricity and has a melting point of about ${2000}^{\circ }C$. Therefore, in the Hall-Heroult process, the electrolysis of aluminium oxide($A{l}_2{O}_3$) is carried out in the bath of cryolite and fluorspar. Cryolite has the chemical name as Sodium aluminium fluoride and chemical formula as ($N{a}_{3}Al{F}_6$). Fluorspar is calcium fluoride having a chemical formula as $Ca{F}_2$. When fused cryolite and fluorspar are added to the aluminium oxide then the melting point of the mixture is decreased to ${900}^{\circ }C$and it also increases the mobility of the fused mixture which in turn increases the conductivity of the solution.

Hence, option A is the correct answer.

Note: Remember that mobility is measurement of how quickly an electron passes through a conductor. As the conductivity of a solution increases then electrons pass through the solution increases which means the mobility increases.