The conductivity of 0.001 `"mol" L^(-1)` solution of `CH_(3)COOH " is " 3.905 xx 10^(-5) "S" cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`.
`"Given" lambda^(@) (H^(+)) = 349.6 "S" cm^(2) "mol"^(-1) " and
" lambda^(0) (CH_(3)COO^(-)) = 40.9 "S" cm^(2) per mol)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate Molar Conductivity (Λ)
The formula for molar conductivity (Λ) is given by:

\[ \Lambda = \frac{\kappa}{C} \]
...