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Ionisation Energy

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Question

First and second ionisation energies of magnesium are 7.646 and 15.035eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into Mg $^{2 +}$ ions present in 12 mg of magnesium vapours is : [Given: 1eV = 96.5 kJ mol $^{- 1}$ ]

A

1.1

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B

1.5

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C

2

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D

0.5

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Solution

The correct option is A 1.1

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Similar questions

Q. First and second ionization energies of Mg are

7.646

15.035

eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into

M g^{2 +}

ions present in 12 mg of magnesium vapours is:

1 e V

= 96.5

k J m o l^{- 1}

]

Q. The first and the second ionisation energies of magnesium are 7.646 eV and 15.035 eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into

M g^{2 +}

ions present in 24 mg of magnesium vapour will be:
(Given

1 e V = 96.5 k J m o l^{- 1}

Q. First and second ionization energies of magnesium are

6.64 e V

13.36 e V

respectively. The amount of energy in J needed to convert all the atoms of magnesium into

M g^{2 +}

ions present in

12 m g

of magnesium vapour will be?

[G i v e n : 1 e V = 96.5 k J / m o l]

Q. The first and the second ionisation energies of magnesium are 7.646 eV and 15.035 eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into

M g^{2 +}

ions present in 24 mg of magnesium vapour will be:
(Given

1 e V = 96.5 k J m o l^{- 1}

Q. First and second ionization energies of magnesium are

7

15 eV

respectively. The amount of energy in

{kJ mol}^{- 1}

needed to convert all the atoms of Magnesium into

M g^{2 +}

ions present in

12 mg

of magnesium vapours is: (Report your answer upto two decimal)
[Given:

1 eV = 96.5 {kJ mol}^{- 1}]

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