The lowering of vapour pressure in a saturated aq. Solution of salt AB is found to be `0.108` torr. If vapour pressure of pure solvent at the same temperature is 300 torr, find the solubility product of salt AB.

Options

A. `10^(-8)`
B. `10^(-6)`
C. `10^(-4)`
D. `10^(-5)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We are given:
- Lowering of vapor pressure (ΔP) = 0.108 torr
- Vapor pressure of pure solvent (P) = 300 torr

### Step 2: Use the formula for relative lowering of vapor pressure
The formula for the relative lowering of vapor pressure is given by:
\[ \frac{\Delta P}{P} = \frac{n}{N} \]
Where:
- \( \Delta P \) = lowering of vapor pressure
- \( P \) = vapor pressure of pure solvent
- \( n \) = number of moles of solute
- \( N \) = number of moles of solvent

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ \frac{0.108}{300} = \frac{n}{N} \]

### Step 4: Calculate the number of moles of solvent (N)
Assuming we have 1 liter of solution, the number of moles of water (solvent) can be calculated as follows:
- Molar mass of water = 18 g/mol
- Density of water = 1 g/mL, so 1 liter of water = 1000 g

Calculating the number of moles of solvent:
\[ N = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.55 \text{ mol} \]

### Step 5: Rearranging the equation to find n
Now we can rearrange the equation to solve for \( n \):
\[ n = \frac{0.108}{300} \times 55.55 \]

### Step 6: Calculate n
Calculating \( n \):
\[ n \approx \frac{0.108 \times 55.55}{300} \approx 0.02 \text{ mol} \]

### Step 7: Find the molarity of the solute
Since we assumed a total volume of 1 liter, the molarity (M) of the solute is:
\[ M = \frac{n}{1 \text{ L}} = 0.02 \text{ M} \]

### Step 8: Calculate the solubility product (Ksp)
For a salt AB that dissociates into A⁺ and B⁻, the solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [A^+][B^-] = s^2 \]
Where \( s \) is the solubility (which is equal to the molarity in this case):
\[ K_{sp} = (0.02)^2 = 0.0004 = 4 \times 10^{-4} \]

### Final Answer
The solubility product \( K_{sp} \) of salt AB is \( 4 \times 10^{-4} \).

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