Calculate the freezing point of a solution of a non-volatile solute in a unknown solvent of molar mass `30g//"mole"` having mole fraction of solvent equal to `0.8`. Given that latent heat of fusion of solid solvent `=2.7 "kcal mol"^(-1)`, freezing point of solvent `=27^(@)C` and `R=2 "cl mol"^(-1)K^(-1)`.

Explanation

To calculate the freezing point of a solution of a non-volatile solute in an unknown solvent, we will follow these steps:

### Step 1: Calculate the Freezing Point Depression Constant (Kf)

The formula for the freezing point depression constant (Kf) is given by:

\[ K_f = \frac{M \cdot R \cdot T_f^2}{1000 \cdot \Delta H_f} \]

Where:
- \(M\) = molar mass of the solvent = 30 g/mol
- \(R\) = gas constant = 2 cal/(mol·K)
- \(T_f\) = freezing point of the solvent in Kelvin = \(273 + 27 = 300 \, K\)
- \(\Delta H_f\) = latent heat of fusion = 2.7 kcal/mol = 2700 cal/mol

Substituting the values into the formula:

\[ K_f = \frac{30 \cdot 2 \cdot 300^2}{1000 \cdot 2700} \]

Calculating \(300^2\):

\[ 300^2 = 90000 \]

Now substituting this back into the equation:

\[ K_f = \frac{30 \cdot 2 \cdot 90000}{1000 \cdot 2700} \]

\[ K_f = \frac{5400000}{2700000} = 2 \, K \cdot kg/mol \]

### Step 2: Calculate the Molality of the Solution

Given the mole fraction of the solvent \(X_{solvent} = 0.8\), we can find the mole fraction of the solute:

\[ X_{solute} = 1 - X_{solvent} = 1 - 0.8 = 0.2 \]

The mole fraction is defined as:

\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]

Where \(n_{solute}\) and \(n_{solvent}\) are the number of moles of solute and solvent, respectively. Rearranging gives:

\[ 0.2 = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]

From the mole fraction of the solvent:

\[ 0.8 = \frac{n_{solvent}}{n_{solute} + n_{solvent}} \]

This gives us the ratio of moles:

\[ \frac{n_{solute}}{n_{solvent}} = \frac{0.2}{0.8} = \frac{1}{4} \]

Let \(n_{solvent} = \frac{w_{solvent}}{M_{solvent}} = \frac{w_{solvent}}{30}\) (where \(w_{solvent}\) is the weight of the solvent in grams).

Thus, we have:

\[ n_{solute} = \frac{1}{4} n_{solvent} = \frac{1}{4} \cdot \frac{w_{solvent}}{30} \]

### Step 3: Calculate the Molality (m)

Molality is defined as:

\[ m = \frac{n_{solute}}{w_{solvent} \, (in \, kg)} \]

Converting \(w_{solvent}\) to kg (1 kg = 1000 g):

\[ m = \frac{\frac{1}{4} \cdot \frac{w_{solvent}}{30}}{\frac{w_{solvent}}{1000}} = \frac{1000}{4 \cdot 30} = \frac{1000}{120} = \frac{25}{3} \approx 8.33 \, mol/kg \]

### Step 4: Calculate the Depression in Freezing Point (\(\Delta T_f\))

The depression in freezing point is given by:

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

Substituting the values:

\[ \Delta T_f = 2 \cdot \frac{25}{3} = \frac{50}{3} \approx 16.67 \, °C \]

### Step 5: Calculate the New Freezing Point

The new freezing point of the solution is:

\[ T_f = T_{f,solvent} - \Delta T_f \]

Substituting the values:

\[ T_f = 27 - 16.67 \approx 10.33 \, °C \]

### Final Answer

The freezing point of the solution is approximately **10.33 °C**.

---