The equivalent conductance of monobasic acid at infinite dilution is `"348 ohm"^(-1)"cm"^(2)"eq"^(-1)`. If the resistivity of the solution containing 15 g acid (mol. Wt 49) in litre is 18.5 ohm cm. What is the `%` degree of dissocition of acid? lReport your answer up to two decimal places without rounding up.

Answer

To solve the problem of finding the percentage degree of dissociation of a monobasic acid, we can follow these steps:

### Step 1: Calculate Molarity of the Acid Solution
Given:
- Mass of the acid = 15 g
- Molar mass of the acid = 49 g/mol

First, we calculate the number of moles of the acid:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{15 \, \text{g}}{49 \, \text{g/mol}} \approx 0.3061 \, \text{mol} \]

Since the volume of the solution is 1 L, the molarity (M) of the solution is:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume (L)}} = \frac{0.3061 \, \text{mol}}{1 \, \text{L}} = 0.3061 \, \text{M} \]

### Step 2: Calculate Conductivity (κ) of the Solution
Given:
- Resistivity (ρ) = 18.5 ohm cm

The conductivity (κ) is the reciprocal of resistivity:
\[ \kappa = \frac{1}{\rho} = \frac{1}{18.5} \approx 0.05405 \, \text{S/cm} \]

### Step 3: Calculate Normality of the Acid Solution
Since the acid is monobasic, the normality (N) is equal to the molarity (M):
\[ N = 0.3061 \, \text{N} \]

### Step 4: Calculate Equivalent Conductance (λ) of the Solution
Using the formula for equivalent conductance:
\[ \lambda = \frac{\kappa \times 1000}{N} \]
Substituting the values we have:
\[ \lambda = \frac{0.05405 \, \text{S/cm} \times 1000}{0.3061} \approx 176.43 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \]

### Step 5: Calculate Degree of Dissociation (α)
Using the formula for degree of dissociation:
\[ \alpha = \frac{\lambda}{\lambda_0} \]
Where:
- \(\lambda_0\) (equivalent conductance at infinite dilution) = 348 ohm\(^{-1}\) cm\(^2\) eq\(^{-1}\)

Substituting the values:
\[ \alpha = \frac{176.43}{348} \approx 0.5065 \]

### Step 6: Calculate Percentage Degree of Dissociation
To find the percentage degree of dissociation:
\[ \% \text{ degree of dissociation} = \alpha \times 100 = 0.5065 \times 100 \approx 50.65 \]

### Final Answer
The percentage degree of dissociation of the acid is approximately **50.65%**.

---