1.0 g of a monobassic acid HA in 100 g water lowers the freezing point by 0.155 K. IF 0.75 g, of same acid requires 25 mL of N/5 NaOH solution for complete neutralisation then %, degree of ionization of acid is (`K_(f) of H_(2)O = 1.86 K kg "mol"^(-1)`):

Options

A. 0.2
B. 0.25
C. 0.4
D. 0.5

Answer

To solve the problem step by step, we will follow the outlined procedure based on the information provided in the question.

### Step 1: Calculate the molality of the solution
We know that the freezing point depression (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
where:
- \( K_f \) is the cryoscopic constant (1.86 K kg/mol for water),
- \( m \) is the molality of the solution,
- \( i \) is the van 't Hoff factor.

Given that \( \Delta T_f = 0.155 \, \text{K} \), we can rearrange the formula to find molality:
\[ m = \frac{\Delta T_f}{K_f \cdot i} \]

### Step 2: Calculate the number of moles of the acid
The molality \( m \) is defined as:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} \]
Given that we have 1 g of acid (HA) in 100 g of water, the mass of the solvent in kg is:
\[ \text{mass of solvent} = 100 \, \text{g} = 0.1 \, \text{kg} \]
Thus, the molality can also be expressed as:
\[ m = \frac{n}{0.1} \]
where \( n \) is the number of moles of the acid.

### Step 3: Find the molecular weight of the acid
We know that 0.75 g of the acid requires 25 mL of N/5 NaOH for complete neutralization. First, we convert the volume of NaOH to liters:
\[ \text{Volume of NaOH} = 25 \, \text{mL} = 0.025 \, \text{L} \]
The normality of NaOH is \( \frac{1}{5} \, \text{N} \), which means:
\[ \text{Equivalents of NaOH} = \text{Normality} \times \text{Volume} = \frac{1}{5} \times 0.025 = 0.005 \, \text{equivalents} \]
Since the acid is monobasic, the equivalents of the acid will also be 0.005. The number of moles of the acid can be calculated as:
\[ \text{Number of moles of acid} = \frac{\text{Equivalents}}{n_f} = \frac{0.005}{1} = 0.005 \]
Now, we can find the molecular weight (M) of the acid:
\[ \text{Molecular weight} = \frac{\text{mass of acid}}{\text{number of moles}} = \frac{0.75 \, \text{g}}{0.005 \, \text{mol}} = 150 \, \text{g/mol} \]

### Step 4: Calculate the van 't Hoff factor (i)
Now we can substitute the values into the molality equation:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Substituting \( m \):
\[ 0.155 = 1.86 \cdot \left(\frac{n}{0.1}\right) \cdot i \]
Substituting \( n \) with the number of moles calculated:
\[ 0.155 = 1.86 \cdot \left(\frac{1 \, \text{g}}{150 \, \text{g/mol} \cdot 0.1}\right) \cdot i \]
Solving for \( i \):
\[ i = \frac{0.155 \cdot 150 \cdot 0.1}{1.86} \]
Calculating \( i \):
\[ i = \frac{2.325}{1.86} \approx 1.25 \]

### Step 5: Calculate the degree of ionization (α)
For a monobasic acid HA dissociating into H⁺ and A⁻:
\[ \text{At equilibrium:} \quad HA \rightleftharpoons H^+ + A^- \]
Let \( \alpha \) be the degree of ionization. Initially, we have 1 mole of HA, and at equilibrium:
- Moles of HA = \( 1 - \alpha \)
- Moles of H⁺ = \( \alpha \)
- Moles of A⁻ = \( \alpha \)

Thus, the total number of moles at equilibrium is:
\[ 1 + \alpha \]
The van 't Hoff factor \( i \) can be expressed as:
\[ i = \frac{1 + \alpha}{1} \]
Substituting \( i \):
\[ 1.25 = 1 + \alpha \]
Solving for \( \alpha \):
\[ \alpha = 1.25 - 1 = 0.25 \]

### Step 6: Calculate the percent degree of ionization
The percent degree of ionization is given by:
\[ \text{Percent degree of ionization} = \alpha \times 100 = 0.25 \times 100 = 25\% \]

### Final Answer
The percent degree of ionization of the acid is **25%**.