Home
Class 12
CHEMISTRY
The molar conductivity of a 1.5 M soluti...

The molar conductivity of a 1.5 M solution of an electrolyte is found to be `138.9 S cm^(2) mol^(-1)` . Calculate the conductivity of this solution.

Text Solution

AI Generated Solution

To calculate the conductivity of the solution, we can use the relationship between molar conductivity (λm), conductivity (κ), and molarity (C) of the solution. The formula is given by: \[ \lambda_m = \frac{\kappa \times 1000}{C} \] Where: - \(\lambda_m\) = Molar conductivity (in S cm² mol⁻¹) ...
Promotional Banner

Topper's Solved these Questions

  • XII BOARDS

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise SET-II|32 Videos
  • XII BOARDS

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise SET-III|31 Videos
  • SAMPLE PAPER 2019

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise SECTION: D|1 Videos

Similar Questions

Explore conceptually related problems

The conductivity of a 0.01 M solution of acetic acid at 298 k is 1.65xx10^(-4) S cm ^(-1) . Calculate molar conductivity (wedge_(m)) of the solution.

The conductivity of a 0.01 M solution of acetic acid at 298 k is 1.65xx10^(-4) S cm ^(-1) . Calculate molar conductivity (wedge_(m)) of the solution.

The molar conductivity of acetic acid solution at infinite dilution is 390.7 Omega^(-1)cm^(2)mol^(-1) . Calculate the molar conductivity of 0.01M acetic acid solution, given that the dissociation of acetic acid is 1.8xx10^(-5) .

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is (100Omega) . The conductivity of this solution is 1.29sm^(-1) . Resistance of the same cell when filled with 0.2 M of the same solution is 520Omega . The molar conductivity of (0.02)M solution of the electrolyte will be :

The equivalent conductance of a 0.2 n solution of an electrolyte was found to be 200Omega^(-1) cm^(2)eq^(-1) . The cell constant of the cell is 2 cm^(-1) . The resistance of the solution is

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100Omega . The conductivity of this solution is 1.29 S m^(-1) . Resistance of the same cell when filled with 0.2 M of the same solution is 520Omega . The molar conductivity of 0.2 M solution of the electrolyte will be :

A weak monobasic acid is 5% dissociated in 0.01 mol dm^(-3) solution. The limiting molar conductivity at infinite dilution is 4.00xx10^(-2) ohm^(-1) m^(2) mol^(-1) . Calculate the conductivity of a 0.05 mol dm^(-3) solution of the acid.

Molar conductivity of a weak acid HA at infinite dilution is 345.8 cm^(2) mol^(-) calculate molar conductivity of 0.05 M HA solution (alpha=5.8 x 10^(-6))

The molar conductivity of 0.05 M solution of weak acid is 16.6Omega^(-1)cm^(-2)mol^(-1). Its molar conductivity at infinite dilution is 390.7Omega^(-1)cm^(-2)mol^(-1) . The degree of dissociation of weak acid is

At 0.04 M concentration, the molar conductivity of solution of an electrolyte is 5000 Omega^(-1)cm^(2) mol^(-1) while at 0.01 M concentration the value is 5100 Omega^(-1)cm^(2)mol^(-1) . Making necessary assumption (Taking it as strong electrolyte) find the molar conductivity at infinite dilution and write percentage dissociation of strong electrolyte at 0.04 M.