Home
Class 11
CHEMISTRY
Calculate heat of solution of NaCl from ...

Calculate heat of solution of NaCl from the following data: Hydration energy of `Na^(oplus) = -389 kJ mol^(-1)`
Hydration energy of ` Cl^(Θ) = - 382 kJ mol^(-1)` , Lattice energy of NaCl = - 776 kJ `mol^(-1)`

A

1

B

3

C

4

D

5

Text Solution

Verified by Experts

The correct Answer is:
5
Hydration energy NaCl = Hydration energy of `Na^(oplus) + ` Hydration energy of `Cl^(Θ)`
`= -389 kJ mol^(-1) - 382 kJ mol^(-1) = -771 kJ mol^(-1)`
Heat of solution `Delta_("(soln H^-)) =` Hydration energy + latttice energy = [-771 + 776- kj `mol^(-1) = 5kj mol^(-1)`
Promotional Banner

Topper's Solved these Questions

  • S-BLOCK ELEMENTS

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Matching Column Type)|7 Videos

  • S-BLOCK ELEMENTS

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Statement Type)|5 Videos

  • S-BLOCK ELEMENTS

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Multiple Choice Question)|10 Videos

  • REDOX REACTIONS

    BRILLIANT PUBLICATION|Exercise LEVEL -III|50 Videos

  • SOME BASIC CONCAPTS OF CHEMISTRY

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Linked Comprehension Type)|6 Videos

Similar Questions

Explore conceptually related problems

Ionisation energy of F^Ɵ is 320 kJ mol^(-1) . The electron gain enthalpy of fluorine would be

The sucessive ionisation energy values for an element X are given below: a) 1st ionisation energy= 410 kJ mol^(-1) b)2nd ionisation energy= 820 kJ mol^(-1) c) 3rd ionisation energy = 1100 kJ mol^(-1) d) 4th ionisation energy = 1500 kJ mol^(-1) 5th ionisation energy= 3200 kJ mol^(-1) Find out the number of valence electron for the atom, X:

The activation energy of exothermic reaction ArarrB is 80 kJ mol^(-1) . The heat of reaction is 200 kJ mol^(-1) . The activation energy for the reaction BrarrA (in kJ mol^(-1) ) will be