The molar conductivity of a 0.5 mol dm³ solution of AgNO₃ with electrolytic conductivity of 5.76x 10⁻³ S cm⁻³ at 298 k in s cm²mol⁻¹ is

A.0.086 S cm²/mol
B.28.8 S cm²/mol
C.2.88 S cm²/mol
D.11.52 S cm²/mol

Respuesta :

Answer:

D. 11.52 S cm²/mol

Explanation:

The formula to calculate the molar conductivity (Λm) of a solution is:

Λm = κ × 1000/c

where,

  • Λm is molar conductivity.
  • κ is electrolytic conductivity
  • c is concentration of the solution.

We have:

  • κ = 5.76 × 10⁻³ S cm⁻³
  • c = 0.5 mol dm⁻³

Plugging the required values, we have:

Λm = 5.76 × 10⁻³ S cm⁻³ × 1000/0.5 mol dm⁻³

Λm = 5760 × 10⁻³ S cm⁻³/0.5 mol dm⁻³

Λm = 11500 × 10⁻³ S cm²/mol

Λm = 11.52 S cm²/mol.

Hence, the molar conductivity of the solution is 11.52 S cm²/mol.

Therefore, (D) 11.52 S cm²/mol is the required answer.

RELAXING NOICE
Relax