Home
Class 12
CHEMISTRY
A 0.05 M NH(4)OH solution offers the res...

A 0.05 M `NH_(4)OH` solution offers the resistance of 50 ohm to a conductivity cell at 298 K. If the cell constant is `0.50cm^(-1)` and molar conductance of `NH_(4)OH` at infinite dilution is `471.4ohm^(-1)cm^(2)mol^(-1)`, calculate :
Molar conductance

Text Solution

AI Generated Solution

To calculate the molar conductance of the 0.05 M `NH4OH` solution, we will follow these steps: ### Step 1: Calculate Specific Conductance (κ) The specific conductance (κ) can be calculated using the formula: \[ \kappa = \frac{1}{R} \cdot \frac{L}{A} \] ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SAMPLE PAPER 2016

    ICSE|Exercise PART-II SECTION-B QUESTION 5. |5 Videos

  • SAMPLE PAPER 2016

    ICSE|Exercise PART-II SECTION-B QUESTION 6. |4 Videos

  • SAMPLE PAPER 2016

    ICSE|Exercise PART-II SECTION-A QUESTION 3. |5 Videos

  • SAMPLE PAPER 2015

    ICSE|Exercise PART-II SECTION-C QUESTION 10. |7 Videos

  • SAMPLE PAPER 4 (CHEMISTRY)

    ICSE|Exercise QUESTIONS|70 Videos

Similar Questions

Explore conceptually related problems

A 0.05 M NH_(4)OH solution offers the resistance of 50 ohm to a conductivity cell at 298 K. If the cell constant is 0.50cm^(-1) and molar conductance of NH_(4)OH at infinite dilution is 471.4ohm^(-1)cm^(2)mol^(-1) , calculate : Specific conductance

A 0.05 M NH_(4)OH solution offers the resistance of 50 ohm to a conductivity cell at 298 K. If the cell constant is 0.50cm^(-1) and molar conductance of NH_(4)OH at infinite dilution is 471.4ohm^(-1)cm^(2)mol^(-1) , calculate : Degree of dissociation

A 0.05 M NH OH solution offers the resistance of 50 ohms to a conductivity cell at 298 K. If the cell constant is 0.50 cm^(-1) and molar conductance of NH_(4)OH at infinite dilution is 471.4 ohm^(-1) cm^(2) "mol"^(-1) calculate : (i) Specific conductance (ii) Molar conductance (iii) Degree of dissociation.

0.05 "M" NaOH solution offered a resistance of 31.6 ohm in a conductivity cell at 298 K. If the cell constant of the cell is 0.367 cm^(_1) calculate the molar conductivity of the NaOH solution.

A 0.05 M NaOH solution offered a resistance of 31.6 Omega in a conductivity cell at 298 K. If the cell constant of the conductivity cell is 0.367 cm^(-1) , find out the specific and molar conductance of the sodium hydroxide solution.

A 0.05 M NaOH solution offered a resistance of 31.6 ohms in a conductivity cell. If the cell constant of the conductivity cell is 0.378 cm^(-1) , determine the molar conductivity of sodium hydroxide solution at this temperature.

A 0.1 M solution of monobasic acid at specific resistance of r ohms-cm, its molar conductivity is

At 298K the resistance of a 0.5N NaOH solution is 35.0 ohm. The cell constant is 0.503 cm^(-1) the electrical conductivity of the solution is

Molar conductance of 2 M H_(2)A acid is 10 S cm^(2) mol^(-1) Molar conductance of H_(2)A at infinite dilution is 400 S cm^(2) mol^(-1) which statement is/are correct?

The conductivity of 0*2M " KCl solution is " 3 Xx 10^(-2) ohm^(-1) cm^(-1) . Calculate its molar conductance.

Home
Profile