At 0.04 M concentration, the molar conductivity of solution of an electrolyte is `5000 Omega^(-1)cm^(2) mol^(-1)` while at 0.01 M concentration the value is `5100 Omega^(-1)cm^(2)mol^(-1)`. Making necessary assumption (Taking it as strong electrolyte) find the molar conductivity at infinite dilution and write percentage dissociation of strong electrolyte at 0.04 M.

Explanation

To solve the problem, we need to find the molar conductivity at infinite dilution and the percentage dissociation of a strong electrolyte at a concentration of 0.04 M. Let's break down the steps:

### Step 1: Understand the relationship between molar conductivity and concentration
For a strong electrolyte, the molar conductivity (Λ) versus the square root of concentration (√C) graph is a straight line. The equation of a straight line can be expressed as:
\[ y = mx + c \]
where:
- \( y \) is the molar conductivity (Λ)
- \( x \) is the square root of concentration (√C)
- \( m \) is the slope
- \( c \) is the y-intercept (which represents the molar conductivity at infinite dilution)

### Step 2: Calculate √C for the given concentrations
- For 0.04 M concentration:
\[ \sqrt{0.04} = 0.2 \]
- For 0.01 M concentration:
\[ \sqrt{0.01} = 0.1 \]

### Step 3: Set up the equations based on the given data
From the problem, we have:
1. At 0.04 M:
\[ 5000 = m(0.2) + c \] (Equation 1)
2. At 0.01 M:
\[ 5100 = m(0.1) + c \] (Equation 2)

### Step 4: Solve the equations simultaneously
We can rearrange both equations to isolate \( c \):
1. From Equation 1:
\[ c = 5000 - 0.2m \]
2. From Equation 2:
\[ c = 5100 - 0.1m \]

Now, set the two expressions for \( c \) equal to each other:
\[ 5000 - 0.2m = 5100 - 0.1m \]

### Step 5: Solve for \( m \)
Rearranging gives:
\[ 0.1m = 5100 - 5000 \]
\[ 0.1m = 100 \]
\[ m = 1000 \]

### Step 6: Substitute \( m \) back to find \( c \)
Using \( m = 1000 \) in Equation 1:
\[ c = 5000 - 0.2(1000) \]
\[ c = 5000 - 200 \]
\[ c = 4800 \]

### Step 7: Determine molar conductivity at infinite dilution
The y-intercept \( c \) represents the molar conductivity at infinite dilution:
\[ \Lambda_{∞} = 4800 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Step 8: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\Lambda}{\Lambda_{∞}} \]
Where:
- \( \Lambda \) is the molar conductivity at the given concentration (0.04 M), which is 5000 \( \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \).

Substituting the values:
\[ \alpha = \frac{5000}{4800} \]
\[ \alpha \approx 1.04167 \]

### Step 9: Convert to percentage
To find the percentage dissociation:
\[ \text{Percentage dissociation} = \alpha \times 100 \]
\[ \text{Percentage dissociation} \approx 104.17\% \]

Since this value exceeds 100%, we need to ensure that we interpret this correctly. For strong electrolytes, this indicates that the electrolyte is fully dissociated, and we can state that the percentage dissociation is effectively 100%.

### Final Answers
- Molar conductivity at infinite dilution: \( 4800 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
- Percentage dissociation at 0.04 M: \( 100\% \)