At 25-C- the specific conductance of 0-01 M aqueous solution of acetic acid is 1-63- 10-2 S m-1 and the molar conductance at infinite dilution is 390- 10-4 S m2mol-1- Calculate the degree of dissociation-

Specific conductance, k = 1.63 × 10^ 2 S m^ 1; c = 0.01 mol dm^ 3 = 0.01 ×10^3 mol m^ 3 Λ_m = kc = 1.63× 10^ 2 S m^ 10.01 × 10^3 mol m^ 1 = 16.3 × 10^ 4 S m^2mo ...

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Standard XII

Chemistry

Degree of Dissociation

At 25∘C, the ...

Question

At $25^{\circ} C$ , the specific conductance of $0.01 M$ aqueous solution of acetic acid is $1.63 \times 10^{- 2} {S m}^{- 1}$ and the molar conductance at infinite dilution is $390 \times 10^{- 4} {S m}^{2} {mol}^{- 1}$ . Calculate the degree of dissociation.

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Solution

$Specific conductance, k = 1.63 \times 10^{- 2} {S m}^{- 1}; c = 0.01 {mol dm}^{- 3} = 0.01 \times 10^{3} {mol m}^{- 3}$
$Λ_{m} = \frac{k}{c} = \frac{1.63 \times 10^{- 2} {S m}^{- 1}}{0.01 \times 10^{3} {mol m}^{- 1}} = 16.3 \times 10^{- 4} {S m}^{2} {mol}^{- 1}$
The degree of dissociation is given by:
$α = \frac{Λ_{m}}{Λ_{m}^{\circ}} = \frac{16.3 \times 10^{- 4} {S m}^{2} {mol}^{- 1}}{390 \times 10^{- 4} {S m}^{2} {mol}^{- 1}} = 0.041$

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At 25∘C, the specific conductance of 0.01 M aqueous solution of acetic acid is 1.63×10−2 S m−1 and the molar conductance at infinite dilution is 390×10−4 S m2mol−1. Calculate the degree of dissociation.

Specific conductance, k=1.63×10−2 S m−1; c=0.01 mol dm−3=0.01×103 mol m−3 Λm=kc=1.63×10−2 S m−10.01×103 mol m−1=16.3×10−4 S m2mol−1 The degree of dissociation is given by: α=ΛmΛ∘m=16.3×10−4 S m2mol−1390×10−4 S m2mol−1=0.041

At $25^{\circ} C$ , the specific conductance of $0.01 M$ aqueous solution of acetic acid is $1.63 \times 10^{- 2} {S m}^{- 1}$ and the molar conductance at infinite dilution is $390 \times 10^{- 4} {S m}^{2} {mol}^{- 1}$ . Calculate the degree of dissociation.