Home
Class 10
CHEMISTRY
One mole of magnesium in the vapor state...

One mole of magnesium in the vapor state absored `1200 kJ mol^(-1)` of enegry. If the first and second ionization energies of `Mg` are `750` and `1450 kJ mol^(-1)`, respectively, the final composition of the mixture is

A

` 69 % Mg^(+), 31% Mg^(2+)`

B

`59% Mg^(+), 41% Mg^(2+)`

C

`49% Mg^(+), 51% Mg^(2+)`

D

`29% Mg^(+), 71% Mg^(2+)`

Text Solution

Verified by Experts

Promotional Banner

Similar Questions

Explore conceptually related problems

First, second & third ionization energies are 737, 1045 & 7733 KJ//mol respectively. The element can be :

Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol^(–1) respectively are:

Calculate the energy required to convert all the atoms of M (atomic number : 12) to M^(2+) ions present in 12 mg of metal vapours. First and second ionization enthalpies of M are 737.77 and 1450.73 kJ " mol"^(-1) respectively.

The heat of combustion of carbon and monoxide are -394 and -285 KJ mol^(-1) respectively. The heat of formation of CO in KJ mol^(-1) is :-