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JEE Main Chemistry question

ChemistryClassification of Elements and Periodicity in PropertiesJEE Main

Ist and IInd IE of Mg are 7.646 and 15.035 eV respectively. The amount of energy (in units of kJ/mol) needed to convert all the atoms of magnesium into ions present in 12 mg of magnesium vapours is [Given, ]

Answer: (C)

Solution

Total energy required to convert one Mg atom into is IE_1 + IE_2 = 7.646 + 15.035 = 22.681\ eV = 22.681 \times 96.5 = 2188.7\ kJ\ mol^{-1}. Moles of Mg in 12 mg $=

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Question ID VK-133310