Home
Class 12
CHEMISTRY
The freezing point of 0.08 molal NaHSO(4...

The freezing point of `0.08 molal NaHSO_(4)` is `-0.345^(@)C`. Calculate the percentage of `HSO_(4)+O` ions that transfers a proton to water. Assume `100%` ionization of `NaHSO_(4)` and `K_(t)` for `H_(2)O = 1.86 K molality^(-1)`.

Text Solution

Verified by Experts

The correct Answer is:
`31.85%` ;
Promotional Banner

Topper's Solved these Questions

  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 3B Objectice Problems|1 Videos

  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 9 Advanced Numerical|12 Videos

  • CHEMICAL KINETICS

    P BAHADUR|Exercise Exercise 9|13 Videos

  • ELECTROCHEMISTRY

    P BAHADUR|Exercise Exercise (9) ADVANCED NUMERICAL PROBLEMS|36 Videos

Similar Questions

Explore conceptually related problems

The freezing point of a 0.08 molal solution of NaHSO^(4) is -0.372^(@)C . Calculate the dissociation constant for the reaction. K_(f) for water = 1.86 K m^(-1)

The freezing point of 0.2 molal K_(2)SO_(4) is -1.1^(@)C . Calculate van't Hoff factor and percentage degree of dissociation of K_(2)SO_(4).K_(f) for water is 1.86^(@)

The freezing point of a 0.08 molal aqueous solutions of NaHSO_(4) is 0.372^(@)C . Calculate the dissociation constant for the reaction HSO_(4)^(-)n hArr H^(+)+SO_(4)^(-2) Assuming no hydrolysis ( k_(t) of water = 1.86 k/kg mole)

Depression in freezing point of 0.1 molal solution of HF is -0.201^(@)C . Calculate percentage degree of dissociation of HF. (K_(f)=1.86 K kg mol^(-1)) .