A very diluted saturated solution of a sparingly soluble salt `X_(3)Y_(4)` has a vapour pressure of 20 mm Hg temperature T,while pure water exerts a pressure of 20.0126 mm Hg at the same temperature . Calculate molality (m)at temperature T :

Options

A. `6.3xx10^(-4)`
B. `3.5xx10^(-2)`
C. `5xx10^(-3)`
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have a sparingly soluble salt \(X_3Y_4\) that forms a very diluted saturated solution. The vapor pressure of this solution is given as \(20 \, \text{mm Hg}\) and the vapor pressure of pure water at the same temperature is \(20.0126 \, \text{mm Hg}\).

### Step 2: Apply Raoult's Law
According to Raoult's law, the relative lowering of vapor pressure is given by:

\[ \frac{P_0 - P}{P_0} = \chi_{solute} \]

Where:
- \(P_0\) = vapor pressure of pure solvent (water)
- \(P\) = vapor pressure of the solution
- \(\chi_{solute}\) = mole fraction of the solute

### Step 3: Calculate the Relative Lowering of Vapor Pressure
Substituting the values we have:

\[ P_0 = 20.0126 \, \text{mm Hg}, \quad P = 20 \, \text{mm Hg} \]

Now, calculate the relative lowering:

\[ \frac{20.0126 - 20}{20.0126} = \frac{0.0126}{20.0126} \approx 0.000629 \]

### Step 4: Determine the Van't Hoff Factor
The salt \(X_3Y_4\) dissociates into ions as follows:

\[ X_3Y_4 \rightarrow 3X^{4+} + 4Y^{3-} \]

This means the total number of ions produced is \(3 + 4 = 7\). Therefore, the Van't Hoff factor \(i = 7\).

### Step 5: Relate Mole Fraction to Molality
The mole fraction of the solute can also be expressed in terms of molality (\(m\)):

\[ \chi_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \approx \frac{m \times \frac{1000}{M_{solvent}}}{m \times \frac{1000}{M_{solvent}} + 1} \]

For very dilute solutions, we can approximate \(n_{solute} + n_{solvent} \approx n_{solvent}\).

### Step 6: Substitute Values into the Equation
Using the approximation and substituting into the equation:

\[ \frac{0.0126}{20.0126} = i \cdot m \cdot \frac{18}{1000} \]

Substituting \(i = 7\):

\[ 0.000629 = 7 \cdot m \cdot \frac{18}{1000} \]

### Step 7: Solve for Molality
Rearranging gives:

\[ m = \frac{0.000629 \cdot 1000}{7 \cdot 18} \]

Calculating the right-hand side:

\[ m = \frac{0.629}{126} \approx 0.0049 \, \text{mol/kg} \]

### Step 8: Final Result
Thus, the molality \(m\) of the solution at temperature \(T\) is approximately:

\[ m \approx 0.005 \, \text{mol/kg} \quad \text{or} \quad 5 \times 10^{-3} \, \text{mol/kg} \]

### Conclusion
The molality of the solution is \(5 \times 10^{-3} \, \text{mol/kg}\).

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