A) \[A{{l}_{2}}{{O}_{3}}+HF+NaAl{{F}_{4}}\]
B) \[A{{l}_{2}}{{O}_{3}}+Ca{{F}_{2}}+NaAl{{F}_{4}}\]
C) \[A{{l}_{2}}{{O}_{3}}+N{{a}_{3}}Al{{F}_{6}}+Ca{{F}_{2}}\]
D) \[A{{l}_{2}}{{O}_{3}}+KF+N{{a}_{3}}Al{{F}_{6}}\]
Correct Answer: C
Solution :
Alumina, \[A{{l}_{2}}{{O}_{3}},\], is a bad conductor of electricity and has very high melting point, so before subjecting to electrolysis, it is mixed with fluorspar \[(Ca{{F}_{2}})\] and cryolite \[(N{{a}_{3}}Al{{F}_{6}})\], which lower its melting point and make it more conducting.
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