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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have:
- Mass of non-volatile solute (W) = 1 g
- Mass of solvent A = 100 g
- Mass of solvent B = 100 g
- \( K_b \) of solvent A to solvent B = 1:5

### Step 2: Write the formula for boiling point elevation
The change in boiling point (\( \Delta T_b \)) is given by the formula:
\[ \Delta T_b = K_b \times m \]
where \( m \) is the molality of the solution.

### Step 3: Calculate molality
Molality (\( m \)) is defined as:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]
The number of moles of solute can be calculated as:
\[ \text{Number of moles} = \frac{W}{M} \]
where \( M \) is the molecular weight of the solute.

Since the mass of the solvent is the same for both A and B (100 g = 0.1 kg), we can express molality as:
\[ m = \frac{1 \text{ g}}{M \text{ g/mol}} \div 0.1 \text{ kg} = \frac{10}{M} \]

### Step 4: Calculate \( \Delta T_b \) for both solvents
For solvent A:
\[ \Delta T_{bA} = K_{bA} \times m = K_{bA} \times \frac{10}{M} \]

For solvent B:
\[ \Delta T_{bB} = K_{bB} \times m = K_{bB} \times \frac{10}{M} \]

### Step 5: Find the ratio \( \frac{\Delta T_{bA}}{\Delta T_{bB}} \)
Now, we can find the ratio of the boiling point elevations:
\[ \frac{\Delta T_{bA}}{\Delta T_{bB}} = \frac{K_{bA} \times \frac{10}{M}}{K_{bB} \times \frac{10}{M}} = \frac{K_{bA}}{K_{bB}} \]

### Step 6: Substitute the given ratio of \( K_b \)
Given that \( \frac{K_{bA}}{K_{bB}} = \frac{1}{5} \):
\[ \frac{\Delta T_{bA}}{\Delta T_{bB}} = \frac{1}{5} \]

### Final Answer
Thus, the final result is:
\[ \frac{\Delta T_{bA}}{\Delta T_{bB}} = \frac{1}{5} \]

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