• 072920 77839
  • info@smartachievers.in
Menu
Back
  • JEE ADVANCE
  • JEE MAIN
  • NEET
  • BITSAT
  • CBSE BOARD
  • UP BOARD
  • CUET
JEE ADVANCE
  • CLASS XI - XII
  • CRASH COURSE
CRASH COURSE
JEE MAIN
  • CLASS XI - XII
  • CRASH COURSE
CRASH COURSE
NEET
  • CLASS XI - XII
  • CRASH COURSE
BITSAT
  • CLASS XI - XII
CBSE BOARD
  • CLASS IX
  • CLASS X
  • CLASS XI
  • CLASS XII
  • CLASS VI
  • CLASS VII
  • CLASS VIII
UP BOARD
  • CLASS IX
  • CLASS X
  • CLASS XI
  • CLASS XII
CUET
  • CRASH COURSE
Courses


Solution


Q.

Consider the given data:

(a)  HCl(g)+10 H2O(l) ⟶ HCl.10H2O    ΔH=-69.01 kJ mol-1

(b)  HCl(g)+40 H2O(l) ⟶ HCl.40H2O    ΔH=-72.79 kJ mol-1

Choose the correct statement:                              [2025]
1 Dissolution of gas in water is an endothermic process.  
2 The heat of solution depends on the amount of solvent.  
3 The heat of dilution for the HCl (HCl.10H2O to HCl.40H2O) is 3.78 kJ mol-1.  
4 The heat of formation of HCl solution is represented by both (a) and (b).  

Ans.

(2)

As heat is evolved during dissolution of HCl(g), it is an exothermic process. As per data given in two equations of question, amount of heat evolved is different for different amount of solvent, hence heat of solution depends upon amount of solvent.

HCl(g)+40H2O(l)→HCl.40H2O,  ΔH1=-72.79 kJ mol-1

HCl(g)+10H2O(l)→HCl.10H2O,  ΔH2=-69.01 kJ mol-1

I-II

HCl.10H2O+30H2O(l)→HCl.40H2O,  ΔH=ΔH1-ΔH2

ΔH=ΔH1-ΔH2

=-72.79-(-69.01)=-3.78 kJ/mol

Neither of the two given equations represent heat of formation of HCl solution, as for heat of formation, reactants must be elements.