If the molar conductivities at infinite dilution at 293K for aqueous hydrochloric acid, sodium acetate and sodium chloride solution are 383.5, 78.4 and 102.0 S `cm^(2)` respectively, calculate themolar conductivity or acetic acid at this temperature and dilution. if the molar conductivity of acetic acid at some other dilution is 100.0 S `cm^(2)` at 293K, calculate the degree of ionization of acetic acid at this dilution.

Answer

To solve the problem step by step, we will use Kohlrausch's law of independent migration of ions. Let's break down the solution:

### Step 1: Calculate the Molar Conductivity of Acetic Acid at Infinite Dilution

According to Kohlrausch's law, the molar conductivity at infinite dilution of a weak electrolyte can be calculated using the molar conductivities of its constituent ions.

The reaction for acetic acid (CH₃COOH) dissociating into ions is:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

We can express the molar conductivity of acetic acid at infinite dilution (Λ°_CH₃COOH) in terms of the molar conductivities of hydrochloric acid (Λ°_HCl), sodium acetate (Λ°_CH₃COONa), and sodium chloride (Λ°_NaCl) as follows:

\[ \Lambda°_{CH₃COOH} = \Lambda°_{HCl} + \Lambda°_{CH₃COONa} - \Lambda°_{NaCl} \]

### Step 2: Substitute the Given Values

Now, we substitute the values provided in the question:

- Λ°_HCl = 383.5 S cm²/mol
- Λ°_CH₃COONa = 78.4 S cm²/mol
- Λ°_NaCl = 102.0 S cm²/mol

Substituting these values into the equation:

\[ \Lambda°_{CH₃COOH} = 383.5 + 78.4 - 102.0 \]

### Step 3: Perform the Calculation

Now, we perform the arithmetic:

\[ \Lambda°_{CH₃COOH} = 383.5 + 78.4 - 102.0 = 359.9 \, \text{S cm}^2/\text{mol} \]

### Step 4: Calculate the Degree of Ionization of Acetic Acid

Now, we need to find the degree of ionization (α) of acetic acid when its molar conductivity at a certain dilution is given as 100.0 S cm²/mol.

The formula for the degree of ionization is:

\[ \alpha = \frac{\Lambda}{\Lambda°_{CH₃COOH}} \]

Where:
- Λ = 100.0 S cm²/mol (molar conductivity at the given dilution)
- Λ°_CH₃COOH = 359.9 S cm²/mol (molar conductivity at infinite dilution)

### Step 5: Substitute the Values

Substituting the values into the formula:

\[ \alpha = \frac{100.0}{359.9} \]

### Step 6: Perform the Calculation

Now, we perform the calculation:

\[ \alpha = 0.2777 \approx 0.278 \]

### Step 7: Convert to Percentage

To express the degree of ionization as a percentage, multiply by 100:

\[ \text{Degree of ionization} = \alpha \times 100 = 0.278 \times 100 = 27.8\% \]

### Final Answers:
1. The molar conductivity of acetic acid at infinite dilution is **359.9 S cm²/mol**.
2. The degree of ionization of acetic acid at the given dilution is **27.8%**.