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The commercial method of preparation of potassium by reduction of molten $KCl$ with metallic sodium at 850° C is based on the fact that:

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sodium has less affinity to chloride ions in the presence of potassium ion

potassium is solid and sodium distils off at 850° C

potassium being more volatile and distils off thus shifting the reaction forward

sodium is more reactive than potassium at 850° C

answer is B.

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Detailed Solution

The reaction is,

$KCl (l) + Na (l) \overset{1123 K (850^{\circ} C)}{⟶} NaCl (l) + K (g)$

Only potassium is in the vapour state at the reaction temperature, allowing it to escape from the reaction mixture.

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