Question

One mole of magnesium in the vapour state absorbed $1200kJmo{l^{ - 1}}$ energy . If the first and second ionisation energies of $Mg$ are $750kJmo{l^{ - 1}}$ and $1450kJmo{l^{ - 1}}$ respectively , the final composition of the mixture is :
A.$86\% Mg + 14\% M{g^{2 + }}$

B.$69\% M{g^ + } + 31\% M{g^{2 + }}$

C.$14\% M{g^ + } + 86\% M{g^{2 + }}$

D.$31\% M{g^ + } + 69\% M{g^{2 + }}$

Answer 1

Hint:The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so as to convert it into a gaseous cation is called its ionization enthalpy . The ionization enthalpy required to remove the second electron from the ion is called the second ionization enthalpy .

Complete step by step answer:
It is given that one mole of magnesium in the vapour state absorbs $1200kJmo{l^{ - 1}}$ of energy .
Since all the values are given for one mole , we don't have to convert anything .
First $Mg$ will ionise to $M{g^ + }$ , the energy absorbed in this ionisation is $750kJmo{l^{ - 1}}$
So , this much energy is used to ionise $Mg$ to $M{g^ + }$
Therefore the energy left unused is
 $1200 - 750 = 450kJmo{l^{ - 1}}$

Now , $450kJmo{l^{ - 1}}$ of energy will be used to convert $M{g^ + }$ to $M{g^{ + 2}}$ .
The number of moles converted to $M{g^{ + 2}}$ can be calculated by dividing the energy used in this process by the second ionisation enthalpy of magnesium .
number of moles converted to $M{g^{ + 2}}$ = $\dfrac{{energy{\text{ }}used{\text{ }}to{\text{ }}convert\;\;M{g^ + }to\;M{g^{ + 2}}\;\;}}{{second{\text{ }}ionisation{\text{ }}enthalpy{\text{ }}of{\text{ }}magnesium}}$

On substituting the values in above equation we get ,

Number of moles converted to $M{g^{ + 2}}$ = $\dfrac{{450}}{{1450}} = 0.3103$

Therefore , number of moles of $M{g^ + }$ = $1 - 0.3103 = 0.69$

Now to find the percentages we will simply multiply the respective moles with 100

Percentage of $M{g^{2 + }}$ = $0.3103 \times 100 = 31\% $ (approximately)

Percentage of $M{g^ + }$ = $0.69 \times 100 = 69\% $

Hence the mixture contains $69\% M{g^ + }$ and $31\% M{g^{2 + }}$
So , option B is correct .


Note: Since in this question, all values were given for 1 mole of a substance only. We didn't have to convert anything but if we are given some other quantity of magnesium gas, we will have to convert the values accordingly .