What is the apparent molecular mass of a weak acid HA (molecular mass= 30 g`//`mole) in an aqueous solution, which shows elevation in boiling point by `0.0156^(@)` C.
Given : `K_(a)(HA)=10^(-2),K_(b)(H_(2)O)=0.52 K kg mol^(-1)`

Answer

To find the apparent molecular mass of the weak acid HA in an aqueous solution, we will follow these steps:

### Step 1: Understand the given data
- Molecular mass of HA = 30 g/mol
- Elevation in boiling point (ΔTb) = 0.0156 °C
- Dissociation constant of HA (Ka) = 10^(-2)
- Molar elevation constant (Kb) = 0.52 K kg/mol

### Step 2: Calculate the molality (m) using the boiling point elevation formula
The formula for boiling point elevation is:
\[ \Delta T_b = K_b \cdot m \]
Rearranging gives us:
\[ m = \frac{\Delta T_b}{K_b} \]
Substituting the values:
\[ m = \frac{0.0156}{0.52} \approx 0.03 \text{ mol/kg} \]

### Step 3: Set up the dissociation of the weak acid HA
The dissociation of HA can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
Let the initial concentration of HA be \(C\) and the amount that dissociates be \(X\). After dissociation:
- Concentration of \(H^+\) = \(X\)
- Concentration of \(A^-\) = \(X\)
- Concentration of undissociated HA = \(C - X\)

### Step 4: Total concentration after dissociation
The total concentration after dissociation is:
\[ C + X \]
From the previous step, we know:
\[ C + X = 0.03 \text{ mol/kg} \]

### Step 5: Use the dissociation constant (Ka)
The expression for the dissociation constant is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{X \cdot X}{C - X} = \frac{X^2}{C - X} \]
Given \(K_a = 10^{-2} = 0.01\), we can set up the equation:
\[ 0.01 = \frac{X^2}{C - X} \]

### Step 6: Substitute \(C\) in terms of \(X\)
From \(C + X = 0.03\), we can express \(C\) as:
\[ C = 0.03 - X \]
Substituting this into the Ka equation:
\[ 0.01 = \frac{X^2}{0.03 - X} \]
Cross-multiplying gives:
\[ 0.01(0.03 - X) = X^2 \]
Expanding and rearranging:
\[ 0.0003 - 0.01X = X^2 \]
\[ X^2 + 0.01X - 0.0003 = 0 \]

### Step 7: Solve the quadratic equation for \(X\)
Using the quadratic formula:
\[ X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = 0.01\), and \(c = -0.0003\):
\[ X = \frac{-0.01 \pm \sqrt{(0.01)^2 - 4 \cdot 1 \cdot (-0.0003)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ X = \frac{-0.01 \pm \sqrt{0.0001 + 0.0012}}{2} \]
\[ X = \frac{-0.01 \pm \sqrt{0.0013}}{2} \]
Calculating the square root:
\[ X \approx \frac{-0.01 \pm 0.036}{2} \]
Taking the positive root:
\[ X \approx \frac{0.026}{2} \approx 0.013 \text{ mol/kg} \]

### Step 8: Calculate \(C\)
Using \(C + X = 0.03\):
\[ C = 0.03 - 0.013 = 0.017 \text{ mol/kg} \]

### Step 9: Calculate the apparent molecular mass
The apparent molecular mass (M') can be calculated using:
\[ M' = \frac{\text{Molecular mass of HA}}{C + X} \]
Substituting the values:
\[ M' = \frac{30 \text{ g/mol}}{0.03} = 20 \text{ g/mol} \]

### Final Answer
The apparent molecular mass of the weak acid HA in the solution is **20 g/mol**.