Calculate molar conductivity of 0.15 M solution of KCl at 298 K if its conductivity is 0.0152 S cm -1 :

Question

Calculate molar conductivity of 0.15 M solution of KCl at 298 K if its conductivity is 0.0152 S cm -1 : (A) 124 Ω-1 cm 2 mol -1 (B) 204 Ω-1 cm 2 mol -1 (C) 101 Ω-1 cm 2 mol -1 (D) 300 Ω-1 cm 2 mol -1

Tardigrade Admin · Accepted Answer

Lambda m =( K × 1000/ M )=(0.0152 × 1000/0.15) =101 Ω-1 cm 2 mol -1

Q. Calculate molar conductivity of $0.15 M$ solution of $K Cl$ at $298 K$ if its conductivity is $0.0152 S c m^{- 1}$ :

$124 Ω^{- 1} c m^{2} m o l^{- 1}$

$204 Ω^{- 1} c m^{2} m o l^{- 1}$

$101 Ω^{- 1} c m^{2} m o l^{- 1}$

$300 Ω^{- 1} c m^{2} m o l^{- 1}$

$Λ_{m} = \frac{K \times 1000}{M} = \frac{0.0152 \times 1000}{0.15}$

$= 101 Ω^{- 1} c m^{2} m o l^{- 1}$

Q. Calculate molar conductivity of 0.15M solution of KCl at 298K if its conductivity is 0.0152Scm−1 :

124Ω−1cm2mol−1

204Ω−1cm2mol−1

101Ω−1cm2mol−1

300Ω−1cm2mol−1