k=4.95 × 10-5 S cm -1 for a 0.001 M solution. The reciprocal of the degree of dissociation of acetic acid, if Lambda m ° for acetic acid is 400 S cm -2 mol -1 is:

Question

k=4.95 × 10-5 S cm -1 for a 0.001 M solution. The reciprocal of the degree of dissociation of acetic acid, if Lambda m ° for acetic acid is 400 S cm -2 mol -1 is: (A) 7 (B) 8 (C) 9 (D) 10

Tardigrade Admin · Accepted Answer

kappa=4.95 × 10-5 S cm -1 ; α=( Lambda m / Lambda m 0) =1000 × ( kappa/C)=1000 × (4.95 × 10-5/0.001)=49.5 S cm -1 ⇒ α=(49.5/400) ⇒ (1/α)=(400/49.5) ≈ 8

Q. $k = 4.95 \times 1 0^{- 5} S c m^{- 1}$ for a $0.001 M$ solution. The reciprocal of the degree of dissociation of acetic acid, if $Λ_{m}^{\circ}$ for acetic acid is $400 S c m^{- 2} m o l^{- 1}$ is:

7

8

9

10

$κ = 4.95 \times 1 0^{- 5} S c m^{- 1}; α = \frac{Λ _{m}}{Λ _{m}^{0}}$
$= 1000 \times \frac{κ}{C} = 1000 \times \frac{4.95 \times 1 0 ^{- 5}}{0.001} = 49.5 S c m^{- 1}$
$\Rightarrow α = \frac{49.5}{400} \Rightarrow \frac{1}{α} = \frac{400}{49.5} \approx 8$

Q. k=4.95×10−5Scm−1 for a 0.001M solution. The reciprocal of the degree of dissociation of acetic acid, if Λm​∘ for acetic acid is 400Scm−2mol−1 is:

7

8

9

10

κ=4.95×10−5Scm−1;α=Λm0​Λm​​ =1000×Cκ​=1000×0.0014.95×10−5​=49.5Scm−1 ⇒α=40049.5​⇒α1​=49.5400​≈8

Q. $k = 4.95 \times 1 0^{- 5} S c m^{- 1}$ for a $0.001 M$ solution. The reciprocal of the degree of dissociation of acetic acid, if $Λ_{m}^{\circ}$ for acetic acid is $400 S c m^{- 2} m o l^{- 1}$ is:

$κ = 4.95 \times 1 0^{- 5} S c m^{- 1}; α = \frac{Λ _{m}}{Λ _{m}^{0}}$
$= 1000 \times \frac{κ}{C} = 1000 \times \frac{4.95 \times 1 0 ^{- 5}}{0.001} = 49.5 S c m^{- 1}$
$\Rightarrow α = \frac{49.5}{400} \Rightarrow \frac{1}{α} = \frac{400}{49.5} \approx 8$