1.0 g of a monobasic 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 approach in the video transcript.

### Step 1: Calculate the Molecular Weight of the Acid (HA)

We can use the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot \frac{m \cdot 1000}{M \cdot W} \]

Where:
- \(\Delta T_f\) = change in freezing point (0.155 K)
- \(K_f\) = freezing point depression constant of water (1.86 K kg/mol)
- \(m\) = mass of the solute (1.0 g)
- \(M\) = molecular weight of the solute (unknown)
- \(W\) = mass of the solvent (100 g)

Rearranging the formula to find \(M\):

\[ M = K_f \cdot \frac{m \cdot 1000}{\Delta T_f \cdot W} \]

Substituting the known values:

\[ M = 1.86 \cdot \frac{1.0 \cdot 1000}{0.155 \cdot 100} \]

Calculating the right-hand side:

\[ M = 1.86 \cdot \frac{1000}{15.5} \approx 120.65 \text{ g/mol} \]

### Step 2: Calculate the Equivalent Weight of the Acid

Given that 0.75 g of the acid requires 25 mL of N/5 NaOH for neutralization, we can find the equivalent weight.

1. Calculate the number of equivalents of NaOH used:
- Normality (N) = N/5 = 0.2 N
- Volume = 25 mL = 0.025 L
- Equivalents of NaOH = Normality × Volume = 0.2 × 0.025 = 0.005 equivalents

Since the acid is monobasic (HA), the equivalents of the acid will be the same:

\[ \text{Equivalents of HA} = 0.005 \]

2. Calculate the equivalent weight of the acid:

\[ \text{Equivalent weight} = \frac{\text{mass of acid}}{\text{equivalents}} = \frac{0.75 \text{ g}}{0.005} = 150 \text{ g/equiv} \]

### Step 3: Calculate the Degree of Ionization

The degree of ionization can be calculated using the formula:

\[ \alpha = \frac{M_{\text{abnormal}} - M_{\text{normal}}}{M_{\text{abnormal}}} \]

Where:
- \(M_{\text{abnormal}} = 150 \text{ g/mol}\) (equivalent weight)
- \(M_{\text{normal}} = 120.65 \text{ g/mol}\) (molecular weight calculated)

Substituting the values:

\[ \alpha = \frac{150 - 120.65}{150} = \frac{29.35}{150} \approx 0.19567 \]

### Step 4: Convert to Percentage Degree of Ionization

To convert the degree of ionization to a percentage:

\[ \text{Percentage degree of ionization} = \alpha \times 100 \approx 0.19567 \times 100 \approx 19.57\% \]

### Final Answer

The percentage degree of ionization of the acid is approximately **19.57%**.

---