The vapour pressure of `0.01m` solution of a weak base `BOH` in water at `20^(@)C` is `17.536mm`. Calculate `K_(b)` for the base. Aq.tension at `20^(@)C=17.54`mm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Vapor pressure of the solution (P_s) = 17.536 mm
- Vapor pressure of pure solvent (P_0) = 17.54 mm
- Molarity of the solution (C) = 0.01 m

### Step 2: Calculate the lowering of vapor pressure
The lowering of vapor pressure (ΔP) can be calculated using the formula:
\[ \Delta P = P_0 - P_s \]
Substituting the values:
\[ \Delta P = 17.54 \, \text{mm} - 17.536 \, \text{mm} = 0.004 \, \text{mm} \]

### Step 3: Calculate the mole fraction of the solute
The mole fraction of the solute (X_solute) can be calculated using Raoult's law:
\[ \Delta P = i \cdot X_{solute} \cdot P_0 \]
Where \(i\) is the van 't Hoff factor. Rearranging gives:
\[ X_{solute} = \frac{\Delta P}{i \cdot P_0} \]
Since we do not know \(i\) yet, we will need to calculate it later.

### Step 4: Calculate the moles of solvent
Assuming we have 1 kg of water (which is approximately 55.56 moles of water):
\[ \text{Moles of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{moles} \]

### Step 5: Calculate the moles of solute
For a 0.01 m solution:
\[ \text{Moles of solute} = 0.01 \, \text{mol} \]

### Step 6: Calculate the mole fraction of the solute
Now, we can calculate the mole fraction of the solute:
\[ X_{solute} = \frac{\text{Moles of solute}}{\text{Moles of solute} + \text{Moles of solvent}} = \frac{0.01}{0.01 + 55.56} \approx \frac{0.01}{55.57} \approx 0.000180 \]

### Step 7: Calculate the van 't Hoff factor (i)
Using the lowering of vapor pressure equation:
\[ \Delta P = i \cdot X_{solute} \cdot P_0 \]
Rearranging gives:
\[ i = \frac{\Delta P}{X_{solute} \cdot P_0} \]
Substituting the values:
\[ i = \frac{0.004}{0.000180 \cdot 17.54} \approx 1.267 \]

### Step 8: Set up the dissociation of the weak base
The weak base \(BOH\) dissociates as follows:
\[ BOH \rightleftharpoons B^+ + OH^- \]
Let \(x\) be the degree of dissociation. At equilibrium:
- Moles of \(B^+\) = \(x\)
- Moles of \(OH^-\) = \(x\)
- Moles of \(BOH\) = \(0.01 - x\)

The van 't Hoff factor \(i\) can be expressed as:
\[ i = 1 + x \]
Setting this equal to our calculated \(i\):
\[ 1 + x = 1.267 \implies x = 0.267 \]

### Step 9: Calculate \(K_b\)
The expression for \(K_b\) is given by:
\[ K_b = \frac{[B^+][OH^-]}{[BOH]} = \frac{x \cdot x}{0.01 - x} = \frac{x^2}{0.01 - x} \]
Substituting the values:
\[ K_b = \frac{(0.267)^2}{0.01 - 0.267} = \frac{0.071289}{-0.257} \approx 9.72 \times 10^{-4} \]

### Final Answer
\[ K_b \approx 9.72 \times 10^{-4} \]