1g of a non-volatile non-electrolyte solute is dissolved in 100g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, `(DeltaT_(b)(A))/(DeltaT_(b)(B))` is

Options

A. `1:0.2`
B. `1:5`
C. `5:1`
D. `10:1`

Explanation

To solve the problem, we need to find the ratio of the elevation in boiling points of two different solvents A and B when a non-volatile non-electrolyte solute is dissolved in them. The ebullioscopic constants of the solvents A and B are given in the ratio of 1:5.

### Step-by-Step Solution:

1. **Understanding the Concept of Boiling Point Elevation**:
The elevation in boiling point (ΔTb) can be calculated using the formula:
\[ \Delta T_b = K_b \cdot m \]
where \( K_b \) is the ebullioscopic constant of the solvent, and \( m \) is the molality of the solution.

2. **Calculating Molality**:
Since we are dissolving 1 g of solute in 100 g of solvent, we need to calculate the molality (m):
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
The mass of solvent A or B is 100 g, which is 0.1 kg.

3. **Finding the Ratio of Elevation in Boiling Points**:
Let’s denote the ebullioscopic constants for solvents A and B as \( K_b(A) \) and \( K_b(B) \), respectively. Given that:
\[ \frac{K_b(A)}{K_b(B)} = \frac{1}{5} \]
The molality for both solutions is the same since the same amount of solute is dissolved in the same mass of solvent (0.1 kg).

4. **Applying the Boiling Point Elevation Formula**:
For solvent A:
\[ \Delta T_b(A) = K_b(A) \cdot m \]
For solvent B:
\[ \Delta T_b(B) = K_b(B) \cdot m \]

5. **Finding the Ratio of Elevation in Boiling Points**:
Now, we can find the ratio of the elevations in boiling points:
\[ \frac{\Delta T_b(A)}{\Delta T_b(B)} = \frac{K_b(A) \cdot m}{K_b(B) \cdot m} = \frac{K_b(A)}{K_b(B)} \]
Substituting the ratio of \( K_b \):
\[ \frac{\Delta T_b(A)}{\Delta T_b(B)} = \frac{1}{5} \]

### Final Answer:
Thus, the ratio of the elevation in boiling points is:
\[ \frac{\Delta T_b(A)}{\Delta T_b(B)} = \frac{1}{5} \]