PREVIOUS EAMCET QUESTIONS with Detailed SOLUTION by VIJCHEM Day 3

 

PREVIOUS EAMCET QUESTIONS with Detailed SOLUTION by #VIJCHEM  #Day-3  #EAMCET

Calculate the energy required to convert all

atoms in 4.8 g of Mg to Mg 2 +

in the vapour

state. IE1and IE2 of Mg are 740 kJ/mol and

1450 kJ/mol respectively. 

[21 Sep. 2020, Shift-I]

(a) + 740 kJ/mol 

(b) − 740 kJ/mol

(c) − 1450 kJ/mol 

(d) + 438 kJ/mol

ANSWER: (d)

DETAILED SOLUTION:

For 1 mole of 

(Mg→ Mg2+ ) 

IE = IE1 + IE2 

=(740+1450)

= 2190 kJ/mol

Number of mole in 4.8g of Mg = 4.8/24 = 0.2 mol

If 

For 1 mole energy required = 2190 kJ/mol

Then

For 0.2 mole energy required = 2190 × 0.2 = 438 kJ

Thus, for 4.8 g of Mg to Mg2 + conversion,

energy required is 438 kJ . Hence, the correct

 *option is (d).*