Home
Class 12
CHEMISTRY
Calculate the heat of neutralization by ...

Calculate the heat of neutralization by mixing `200mL` of `0.1 MH_(2)SO_(4)` and `200mL ` of `0.2 MKOH `if heat generated by the mixing is `2.3 kJ`.

Text Solution

Verified by Experts

Meq. Of `H_(2)SO_(4) = 200 xx 0.1 xx 2 = 40` `(H_(2)SO_(4))` is diabasic
Meq. Of KOH = `200 xx 0.2 = 40`
`therefore 40` Meq. Of `H_(2)SO_(4)` and 40 Meq. Of KOH on mixing gives heat `=(2.3 xx 1000)/40 = 57.5` kJ
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS AND THERMOCHEMISTRY

    FIITJEE|Exercise SOLVED PROBLEM (SUBJECTIVE)|13 Videos

  • THERMODYNAMICS AND THERMOCHEMISTRY

    FIITJEE|Exercise SOLVED PROBLEMS (OBJECTIVE)|28 Videos

  • TEST PAPERS

    FIITJEE|Exercise CHEMISTRY|747 Videos

  • TRANSITION ELEMENTS & COORDINATION COMPOUNDS

    FIITJEE|Exercise MATCHIG LIST TYPE QUESTIONS|1 Videos

Similar Questions

Explore conceptually related problems

The normality of mixture obtained by mixing 100 mL of 0.2 M H_(2)SO_(4) and 200 mL of 0.2 M HCl is

The nature of mixture obtained by mixing 50mL of 0.1M H_(2)SO_(4) and 50mL of 0.1M NaOh is:

The normality of solution obtained by mixing 100 ml of 0.2 M H_(2) SO_(4) with 100 ml of 0.2 M NaOH is

Calculate the pH of solution obtained by mixing 10 ml of 0.1 M HCl and 40 ml of 0.2 M H_(2)SO_(4)

The weight of solute present in 200mL of 0.1M H_(2)SO_(4) :

Calculate the heat of neutralisation from the following data: 200mL of 1M HCI is mixed with 400mL of 0.5M NaOH . The temperature rise in calorimeter was found to be 4.4^(@)C . Water equivalent of calorimeter is 12g and specific heat is 1cal mL^(-1) degree^(-1) for solution.