If the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associating solute in solvent (`K_(b) =xK.kg"mol"^(-1)`) is yK,then the depression in freezing point of solution of same concentration would be
(`K_(f)` )of the solvent = `zK.kg"mol"^(-1)`)

Options

A. `2xy/y`
B. `yz/x`
C. `xz/y`
D. `yz/(2x)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given:
- Elevation in boiling point, \( \Delta T_b = y \, K \)
- Boiling point elevation constant, \( K_b = x \, K \cdot kg \, mol^{-1} \)
- Freezing point depression constant, \( K_f = z \, K \cdot kg \, mol^{-1} \)

### Step 2: Write the Formulas for Boiling Point Elevation and Freezing Point Depression
The formulas for boiling point elevation and freezing point depression are:
- \( \Delta T_b = i \cdot m \cdot K_b \)
- \( \Delta T_f = i \cdot m \cdot K_f \)

Where:
- \( i \) is the van 't Hoff factor (which is 1 for non-electrolytic and non-associating solutes)
- \( m \) is the molality of the solution

### Step 3: Set Up the Ratios
Since the solute is non-electrolytic and non-associating, we can assume \( i = 1 \). Therefore, we can write:
- \( \Delta T_b = m \cdot K_b \)
- \( \Delta T_f = m \cdot K_f \)

Now, we can set up the ratio of the freezing point depression to the boiling point elevation:
\[ \frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b} \]

### Step 4: Substitute the Known Values
Substituting the known values into the ratio:
\[ \frac{\Delta T_f}{y} = \frac{z}{x} \]

### Step 5: Solve for \( \Delta T_f \)
Cross-multiplying gives:
\[ \Delta T_f = \frac{z}{x} \cdot y \]

### Step 6: Final Expression
Thus, the depression in freezing point is given by:
\[ \Delta T_f = \frac{zy}{x} \]

### Conclusion
The depression in freezing point of the solution of the same concentration is \( \Delta T_f = \frac{zy}{x} \).

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