Ratio of `DeltaT_(b)//K_(b) of 6% AB_(2) and 9% A_(2)BAB_(2) and A_(2)B` both are non-electrolytes is 1 mol/kg in both cases. Hence atomic masses of A and B are respetively.

Options

A. `60,90`
B. `40,40`
C. `40,10`
D. `10,40`

Explanation

To solve the question regarding the ratio of `DeltaT_(b)//K_(b)` for the solutions of 6% AB₂ and 9% A₂BAB₂, we need to follow these steps:

### Step 1: Understand the Problem
We are given two solutions:
1. 6% AB₂
2. 9% A₂BAB₂

Both solutions are non-electrolytes and have the same molality (1 mol/kg). We need to find the atomic masses of A and B based on the ratio of their boiling point elevation constants.

### Step 2: Define the Boiling Point Elevation Formula
The boiling point elevation can be expressed using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( \Delta T_b \) = boiling point elevation
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_b \) = ebullioscopic constant of the solvent
- \( m \) = molality of the solution

### Step 3: Calculate the Van 't Hoff Factor (i)
For non-electrolytes, the van 't Hoff factor \( i \) is equal to 1. Thus, for both solutions:
- For 6% AB₂: \( i = 1 \)
- For 9% A₂BAB₂: \( i = 1 \)

### Step 4: Set Up the Ratio
Given that the ratio of \( \Delta T_b / K_b \) for both solutions is equal to 1, we can set up the following equation:
\[ \frac{\Delta T_{b1}}{K_{b1}} = \frac{\Delta T_{b2}}{K_{b2}} \]

### Step 5: Express the Boiling Point Elevation
Using the boiling point elevation formula:
\[ \Delta T_{b1} = K_{b1} \cdot m \]
\[ \Delta T_{b2} = K_{b2} \cdot m \]

Since both solutions have the same molality (1 mol/kg), we can simplify the ratio:
\[ \frac{K_{b1}}{K_{b2}} = \frac{6\%}{9\%} \]

### Step 6: Calculate the Ratio
Now, substituting the percentages:
\[ \frac{K_{b1}}{K_{b2}} = \frac{6}{9} = \frac{2}{3} \]

### Step 7: Relate to Molar Mass
The boiling point elevation constant \( K_b \) is related to the molar mass of the solute. Therefore, we can set up the relationship:
\[ \frac{M_{A}}{M_{B}} = \frac{2}{3} \]

### Step 8: Solve for Atomic Masses
Let the atomic mass of A be \( M_A \) and that of B be \( M_B \). From the ratio, we can express:
\[ M_A = \frac{2}{3} M_B \]

### Step 9: Conclusion
To find the specific atomic masses, we would need additional information or constraints (like the total mass of the solute). However, based on the ratio derived, we can conclude that the atomic masses of A and B are related as \( M_A : M_B = 2 : 3 \).