Molar conductance of 0.1 M acetic acid is 7 ohm-1 cm2 mol-1. If the molar conductance of acetic acid at infinite dilution is 380.8 ohm-1 cm2 mol-1, the value of dissociation constant will be:

Question

Molar conductance of 0.1 M acetic acid is 7 ohm-1 cm2 mol-1. If the molar conductance of acetic acid at infinite dilution is 380.8 ohm-1 cm2 mol-1, the value of dissociation constant will be: (A) 226× 10-5 mol dm-3 (B) 1.66 × 10-3 mol dm-1 (C) 1.66 × 10-2 mol dm-3 (D) 3.38 × 10-5 mol dm-3

Tardigrade Admin · Accepted Answer

Ka=Cα2=0.1 ×((7/380.8))2 =3.38×10-5

Q. Molar conductance of 0.1 M acetic acid is $7 o h m^{- 1} c m^{2} m o l^{- 1}$ . If the molar conductance of acetic acid at infinite dilution is $380.8 o h m^{- 1} c m^{2} m o l^{- 1}$ , the value of dissociation constant will be:

$226 \times 1 0^{- 5} m o l d m^{- 3}$

$1.66 \times 1 0^{- 3} m o l d m^{- 1}$

$1.66 \times 1 0^{- 2} m o l d m^{- 3}$

$3.38 \times 1 0^{- 5} m o l d m^{- 3}$

$K_{a} = C α^{2} = 0.1 \times (\frac{7}{380.8})^{2} = 3.38 \times 1 0^{- 5}$

Q. Molar conductance of 0.1 M acetic acid is 7ohm−1cm2mol−1. If the molar conductance of acetic acid at infinite dilution is 380.8ohm−1cm2mol−1, the value of dissociation constant will be:

226×10−5moldm−3

1.66×10−3moldm−1

1.66×10−2moldm−3

3.38×10−5moldm−3