The molar conductivity of 0.05 M of solution of an electrolyte is 200 Ω-1 cm2 mol-1. The resistance offered by a conductivity cell with cell constant (1/3) cm-1 would be about

Question

The molar conductivity of 0.05 M of solution of an electrolyte is 200 Ω-1 cm2 mol-1. The resistance offered by a conductivity cell with cell constant (1/3) cm-1 would be about (A) 11.11 Ω (B) 22.22 Ω (C) 33.33 Ω (D) 44.44 Ω

Tardigrade Admin · Accepted Answer

k =∧c=(200 S cm2 mol-1)(0.05 ×10-3 mol cm-1) =0.01 S cm-1 R=(1/k)((l/a)) =(1/(0.01 S cm-1))((1/3)cm-1)=33.33 Ω

Q. The molar conductivity of $0.05 M$ of solution of an electrolyte is $200 Ω^{- 1} c m^{2} m o l^{- 1}$ . The resistance offered by a conductivity cell with cell constant $(1/3) c m^{- 1}$ would be about

11.11 $Ω$

22.22 $Ω$

33.33 $Ω$

44.44 $Ω$

$k = \land_{c} = (200 S c m^{2} m o l^{- 1}) (0.05 \times 1 0^{- 3} m o l c m^{- 1})$
$= 0.01 S c m^{- 1}$
$R = \frac{1}{k} (\frac{l}{a}) = \frac{1}{( 0.01 S c m ^{- 1} )} (\frac{1}{3} c m^{- 1}) = 33.33 Ω$

Q. The molar conductivity of 0.05M of solution of an electrolyte is 200Ω−1cm2mol−1. The resistance offered by a conductivity cell with cell constant (1/3)cm−1 would be about

11.11 Ω

22.22 Ω

33.33 Ω

44.44 Ω