The specific conductance of 0.01 M solution of acetic acid was found to be 0.0163 Sm-1 at 25°C . Molar conductance of acetic acid at infinite dilution is 390.7 × 10-4 Sm2 mol-1 at 25° C . What will be the degree of dissociation of CH3COOH ?

Question

The specific conductance of 0.01 M solution of acetic acid was found to be 0.0163 Sm-1 at 25°C . Molar conductance of acetic acid at infinite dilution is 390.7 × 10-4 Sm2 mol-1 at 25° C . What will be the degree of dissociation of CH3COOH ? (A)  0.4072  (B)  0.7402  (C)  0.2720  (D)  0.0472

Tardigrade Admin · Accepted Answer

Degree of dissociation at 25°C α = (μc/μα) = (0.0163 Sm-1/390.7 × 10-4 Sm2) mol-1 = 0.0472

$0.4072$

$0.7402$

$0.2720$

$0.0472$

Degree of dissociation at $2 5^{\circ} C$
$α = \frac{μ ^{c}}{μ ^{α}}$
$= \frac{0.0163 S m ^{- 1}}{390.7 \times 1 0 ^{- 4} S m ^{2}} m o l^{- 1}$
$= 0.0472$

Q. The specific conductance of 0.01M solution of acetic acid was found to be 0.0163Sm−1 at 25∘C . Molar conductance of acetic acid at infinite dilution is 390.7×10−4Sm2mol−1 at 25∘C . What will be the degree of dissociation of CH3​COOH ?

0.4072

0.7402

0.2720

0.0472

Degree of dissociation at 25∘C α=μαμc​ =390.7×10−4Sm20.0163Sm−1​mol−1 =0.0472

$0.4072$

$0.7402$

$0.2720$

$0.0472$

Degree of dissociation at $2 5^{\circ} C$
$α = \frac{μ ^{c}}{μ ^{α}}$
$= \frac{0.0163 S m ^{- 1}}{390.7 \times 1 0 ^{- 4} S m ^{2}} m o l^{- 1}$
$= 0.0472$