Solveeit Logo
Login

Question

Question: One mole of magnesium in the vapour state absorbed \(1200kJmo{l^{ - 1}}\) energy . If the first and ...

One mole of magnesium in the vapour state absorbed 1200kJmol−11200kJmo{l^{ - 1}} energy . If the first and second ionisation energies of MgMg are 750kJmol−1750kJmo{l^{ - 1}} and 1450kJmol−11450kJmo{l^{ - 1}} respectively , the final composition of the mixture is :
A.86%Mg+14%Mg2+86\% Mg + 14\% M{g^{2 + }}

B.69%Mg++31%Mg2+69\% M{g^ + } + 31\% M{g^{2 + }}

C.14%Mg++86%Mg2+14\% M{g^ + } + 86\% M{g^{2 + }}

D.31%Mg++69%Mg2+31\% M{g^ + } + 69\% M{g^{2 + }}

Explanation

Solution

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 1200kJmol−11200kJmo{l^{ - 1}} of energy .
Since all the values are given for one mole , we don't have to convert anything .
First MgMg will ionise to Mg+M{g^ + } , the energy absorbed in this ionisation is 750kJmol−1750kJmo{l^{ - 1}}
So , this much energy is used to ionise MgMg to Mg+M{g^ + }
Therefore the energy left unused is
1200−750=450kJmol−11200 - 750 = 450kJmo{l^{ - 1}}

Now , 450kJmol−1450kJmo{l^{ - 1}} of energy will be used to convert Mg+M{g^ + } to Mg+2M{g^{ + 2}} .
The number of moles converted to Mg+2M{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 Mg+2M{g^{ + 2}} = energy used to convert    Mg+to  Mg+2    second ionisation enthalpy of magnesium\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 Mg+2M{g^{ + 2}} = 4501450=0.3103\dfrac{{450}}{{1450}} = 0.3103

Therefore , number of moles of Mg+M{g^ + } = 1−0.3103=0.691 - 0.3103 = 0.69

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

Percentage of Mg2+M{g^{2 + }} = 0.3103×100=31%0.3103 \times 100 = 31\% (approximately)

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

Hence the mixture contains 69%Mg+69\% M{g^ + } and 31%Mg2+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 .