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Chemistry - Metallurgy Question with Solution | TestHub

ChemistryMetallurgyEllingham diagram & Reduction methods, Refining of metalMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

A metal " $M_{1}$ " present as an impurity in metal " $M_{2}$ ", is refined by liquation. Sulphide of metal " $M_{2}$ " is separated from sulphide of metal " $M_{1}$ " by adding KCN during concentration of sulphide of " $M_{1}$ ". Both " $M_{1}$ " & " $M_{2}$ " predominantly have + 2 oxidation state in their compounds.

The refining process that can be employed for $M_{2}$ is :-

Options:

Answer:

Solution:

The question refers to refining $M_{2}$ , which is likely a typo and should be $H g$ (mercury). Mercury is refined by fractional distillation due to its low boiling point.

Stream:JEESubject:ChemistryTopic:MetallurgySubtopic:Ellingham diagram & Reduction methods, Refining of metal

⏱ 2mℹ️ Source: QB

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