1.0 gram of a monobasic acid HA in 100 gram `H_2O` lower the freezing point by 0.155 K.0.45 gram of same acid require 15 ml of `1/5`M KOH solution for complete neutralisation.If the degree of dissociation of acid is `alpha`, then value of `'20alpha'` is : (`K_f` for `H_2O` =1.86 K.Kg/mole)

Explanation

To solve the problem step by step, we will break down the information given and apply the relevant formulas.

### Step 1: Calculate the molality of the solution
Given:
- Mass of the acid (HA) = 1.0 g
- Mass of water (solvent) = 100 g = 0.1 kg

First, we need to find the molality (m) of the solution. The formula for molality is:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

To find the moles of the acid, we need its molar mass. We will find it later, but for now, let's denote the molar mass of HA as \( M \).

The number of moles of HA is:

\[ \text{moles of HA} = \frac{1.0 \, \text{g}}{M} \]

Thus, the molality (m) becomes:

\[ m = \frac{1.0/M}{0.1} = \frac{10}{M} \, \text{mol/kg} \]

### Step 2: Use the freezing point depression formula
The freezing point depression (\( \Delta T_f \)) is given by:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \( \Delta T_f = 0.155 \, \text{K} \)
- \( K_f = 1.86 \, \text{K kg/mol} \)

Substituting the values we have:

\[ 0.155 = i \cdot 1.86 \cdot \frac{10}{M} \]

Rearranging gives:

\[ i = \frac{0.155 \cdot M}{18.6} \]

### Step 3: Determine the degree of dissociation (\( \alpha \))
For a monobasic acid (HA), it dissociates as follows:

\[ HA \rightleftharpoons H^+ + A^- \]

The van 't Hoff factor \( i \) for this dissociation is given by:

\[ i = 1 + \alpha \]

Where \( \alpha \) is the degree of dissociation. Therefore, we can rewrite:

\[ 1 + \alpha = \frac{0.155 \cdot M}{18.6} \]

### Step 4: Find the molar mass of the acid using neutralization
From the neutralization information:
- 0.45 g of HA requires 15 mL of \( \frac{1}{5} \, M \) KOH.

Calculating the moles of KOH used:

\[ \text{Moles of KOH} = M \cdot V = \frac{1}{5} \cdot 0.015 = 0.003 \, \text{moles} \]

Since HA is monobasic, it reacts in a 1:1 ratio with KOH, so:

\[ \text{Moles of HA} = 0.003 \]

Now, we can find the molar mass \( M \) of HA:

\[ M = \frac{0.45 \, \text{g}}{0.003 \, \text{moles}} = 150 \, \text{g/mol} \]

### Step 5: Substitute back to find \( \alpha \)
Now substituting \( M = 150 \) g/mol back into the equation for \( i \):

\[ i = \frac{0.155 \cdot 150}{18.6} \approx 1.25 \]

Now, substituting into the equation for \( \alpha \):

\[ 1 + \alpha = 1.25 \implies \alpha = 0.25 \]

### Step 6: Calculate \( 20\alpha \)
Finally, we calculate \( 20\alpha \):

\[ 20\alpha = 20 \cdot 0.25 = 5 \]

### Final Answer
Thus, the value of \( 20\alpha \) is **5**.

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