The value of `K_(b) ` for a solvent is X K `kg mol ^(-1) ` A 0.2 m solution of a non-electrolyte in this solvent will boil at ___ .( Given : boiling point of solvent `=A^(@)C`)

Options

A. `(A+X)^(@)C`
B. `(A+(X )/(10))^(@)C`
C. `(A+(X )/(5 ))^@C`
D. ` (A+(X )/(5))K`

Answer

To solve the problem, we need to calculate the boiling point of a 0.2 m solution of a non-electrolyte in a solvent, given that the boiling point elevation constant \( K_b \) for the solvent is \( x \) K kg mol\(^{-1}\) and the boiling point of the solvent is \( A \) °C.

### Step-by-Step Solution:

1. **Understand the Formula for Boiling Point Elevation**:
The boiling point elevation (\( \Delta T_b \)) can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
where \( K_b \) is the boiling point elevation constant and \( m \) is the molality of the solution.

2. **Substitute the Given Values**:
In this case, we know:
- \( K_b = x \) K kg mol\(^{-1}\)
- \( m = 0.2 \) m

Therefore, substituting these values into the formula gives:
\[ \Delta T_b = x \times 0.2 \]

3. **Calculate the Elevation**:
Simplifying the equation:
\[ \Delta T_b = 0.2x \]

4. **Determine the New Boiling Point**:
The new boiling point (\( T_b \)) of the solution can be calculated by adding the boiling point elevation to the original boiling point of the solvent:
\[ T_b = A + \Delta T_b \]
Substituting \( \Delta T_b \) into this equation gives:
\[ T_b = A + 0.2x \]

5. **Final Expression**:
Thus, the boiling point of the 0.2 m solution of the non-electrolyte in the solvent is:
\[ T_b = A + 0.2x \text{ °C} \]

### Final Answer:
The boiling point of the 0.2 m solution of a non-electrolyte in this solvent will boil at \( A + 0.2x \) °C.

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