Calculate solubility (in moles/litre) of a saturated aqueous solution
of `Ag_(3)PO_(4)` if the vapour pressure of the solution becomes 750 torr
at 373 K:

Options

A. 0.08
B. 0.18
C. 0.28
D. 0.38

Answer

To calculate the solubility (in moles/litre) of a saturated aqueous solution of \( \text{Ag}_3\text{PO}_4 \) given that the vapor pressure of the solution is 750 torr at 373 K, we can follow these steps:

### Step 1: Understand the Problem
We are dealing with a saturated solution of \( \text{Ag}_3\text{PO}_4 \) in water at its boiling point (373 K). The vapor pressure of pure water at this temperature is 760 torr, and the vapor pressure of the solution is given as 750 torr.

### Step 2: Calculate the Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure can be calculated using the formula:

\[ \frac{P_0 - P_s}{P_0} = \frac{760 \, \text{torr} - 750 \, \text{torr}}{760 \, \text{torr}} = \frac{10}{760} = \frac{1}{76} \]

### Step 3: Determine the Van 't Hoff Factor (i)
For \( \text{Ag}_3\text{PO}_4 \), it dissociates in water as follows:

\[ \text{Ag}_3\text{PO}_4 (s) \rightarrow 3 \text{Ag}^+ (aq) + \text{PO}_4^{3-} (aq) \]

This means that for each mole of \( \text{Ag}_3\text{PO}_4 \), we get 4 moles of ions (3 moles of \( \text{Ag}^+ \) and 1 mole of \( \text{PO}_4^{3-} \)). Therefore, the van 't Hoff factor \( i = 4 \).

### Step 4: Relate Vapor Pressure Lowering to Molarity
Using Raoult's law, we can relate the lowering of vapor pressure to the molality of the solution:

\[ \frac{P_0 - P_s}{P_0} = \frac{i \cdot n_b}{n_a} \]

Where:
- \( n_b \) = moles of solute (Ag3PO4)
- \( n_a \) = moles of solvent (water)

Assuming 1 kg of water (which is approximately 55.5 moles), we can express the equation as:

\[ \frac{1}{76} = \frac{4 \cdot n_b}{55.5} \]

### Step 5: Solve for \( n_b \)
Rearranging the equation gives:

\[ n_b = \frac{55.5}{4 \cdot 76} \]

Calculating this:

\[ n_b = \frac{55.5}{304} \approx 0.182 mol \]

### Step 6: Calculate Molarity
Since we assumed 1 kg of water, the volume of the solution is approximately 1 L (since the density of water is about 1 kg/L). Therefore, the molarity (which is moles per liter) is:

\[ \text{Molarity} = n_b \approx 0.182 \, \text{mol/L} \]

### Final Answer
The solubility of \( \text{Ag}_3\text{PO}_4 \) in the saturated aqueous solution is approximately **0.182 mol/L**.

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