The molar conductivity of 0.1 M CH3COOH solution is 4 S cm2 mol–1. What is the resistivity of the solution ?2200 Ω cm2500 Ω cm2400 Ω cm2000 Ω cm
Question
The molar conductivity of 0.1 M CH3COOH solution is 4 S cm2 mol–1. What is the resistivity of the solution ?2200 Ω cm2500 Ω cm2400 Ω cm2000 Ω cm
Solution
The resistivity of a solution can be calculated using the formula:
Resistivity (ρ) = 1 / Conductivity
In this case, the molar conductivity is given as 4 S cm^2 mol^-1. However, this is the conductivity per mole, and we have a 0.1 M solution. So, the actual conductivity (κ) of the solution is:
κ = Molar Conductivity * Concentration κ = 4 S cm^2 mol^-1 * 0.1 mol/L κ = 0.4 S/cm
Now, we can calculate the resistivity:
ρ = 1 / κ ρ = 1 / 0.4 S/cm ρ = 2.5 Ω cm
So, the resistivity of the solution is 2.5 Ω cm.
Similar Questions
The cell constant of a conductivity cell is 0.146 cm-1. What is the conductivity of 0.01 M solutionof an electrolyte at 298 K, if the resistance of the cell is 1000 ohm?
The resistivity of a column of 0.1 M KCl solution is 77.51 ohm·cm. The molar conductivity of KCl solution is
Resistance of a conductivity cell (cell constant 129 m–1) filled with 74.5 ppm solution of KCl is 100 Ω (labeled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is 50 Ω (labeled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. ∧1∧2 = x × 10−3. The value of x is______. (Nearest integer)Given, molar mass of KCl is 74.5 g mol–1.
Which of the following solutions will have the highest conductivity at 298 K?(a) 0.01 M HCl solution (b) 0.1 M HCl solution(c) 0.01 M CH 3COOH solution (d) 0.1 M CH 3COOH solution
Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.