`1g` of same non volatile solute is added to `100g` of two different solvents `A` and `B`. `K_(b)` of `A:B=1:5`
Find out `((DeltaT_(b))_(A))/((DeltaT_(b))_(B))`

Options

A. `1:10`
B. `1:5`
C. `1:20`
D. `2:5`

Answer

To solve the problem, we need to find the ratio of the boiling point elevation for two different solvents (A and B) when the same non-volatile solute is added to both.

### Step-by-Step Solution:

1. **Understand the Formula for Boiling Point Elevation**:
The boiling point elevation (\( \Delta T_b \)) is given by 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. **Identify the Given Information**:
- Mass of solute = 1 g
- Mass of solvent A = 100 g
- Mass of solvent B = 100 g
- Ratio of \( K_b \) for solvents A and B = \( K_b(A) : K_b(B) = 1 : 5 \)

3. **Calculate Molality for Solvent A**:
Molality (\( m \)) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
The moles of solute can be calculated as:
\[ \text{moles} = \frac{1 \text{ g}}{M} = \frac{1}{M} \]
where \( M \) is the molar mass of the solute (in grams per mole). The mass of solvent A in kg is:
\[ 100 \text{ g} = 0.1 \text{ kg} \]
Therefore, the molality of A is:
\[ m_A = \frac{1/M}{0.1} = \frac{10}{M} \]

4. **Calculate Molality for Solvent B**:
The mass of solvent B is also 100 g, which is:
\[ 0.1 \text{ kg} \]
Thus, the molality of B is the same as A since the mass of solute and solvent are the same:
\[ m_B = \frac{1/M}{0.1} = \frac{10}{M} \]

5. **Calculate the Ratio of Boiling Point Elevation**:
Now we can express the boiling point elevation for both solvents:
\[ \Delta T_{bA} = K_b(A) \cdot m_A = K_b(A) \cdot \frac{10}{M} \]
\[ \Delta T_{bB} = K_b(B) \cdot m_B = K_b(B) \cdot \frac{10}{M} \]
Taking the ratio of the boiling point elevations:
\[ \frac{\Delta T_{bA}}{\Delta T_{bB}} = \frac{K_b(A) \cdot m_A}{K_b(B) \cdot m_B} = \frac{K_b(A)}{K_b(B)} \cdot \frac{m_A}{m_B} \]

6. **Substituting the Known Values**:
Since \( K_b(A) : K_b(B) = 1 : 5 \), we have:
\[ \frac{K_b(A)}{K_b(B)} = \frac{1}{5} \]
And since \( m_A = m_B \), we have:
\[ \frac{m_A}{m_B} = 1 \]
Therefore:
\[ \frac{\Delta T_{bA}}{\Delta T_{bB}} = \frac{1}{5} \cdot 1 = \frac{1}{5} \]

### Final Answer:
\[ \frac{\Delta T_{bA}}{\Delta T_{bB}} = \frac{1}{5} \]