The elements X and Y form compound having molecular formula `XY_(2) and XY_(4)` (both are non - electrolysis), when dissolved in 20 g benzene, `1 g XY_(2)` lowers the freezing point by `2.3^(@)c` whereas 1 g of `XY_(4)` lowers the freezing point by `1.3^(C)`. Molal depression constant for benzene is 5.1. Thus atomic masses of X and Y respectively are

Options

A. 42.64, 21.10
B. 21.10, 42.64
C. 25.59, 42.64
D. 42.64, 25.69

Answer

To solve the problem, we need to determine the atomic masses of elements X and Y based on the freezing point depression data provided for the compounds XY₂ and XY₄.

### Step-by-Step Solution:

1. **Understanding Freezing Point Depression**:
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \(\Delta T_f\) = depression in freezing point
- \(i\) = van 't Hoff factor (which is 1 for non-electrolytes)
- \(K_f\) = molal depression constant (given as 5.1 for benzene)
- \(m\) = molality of the solution

2. **Calculating Molality for XY₂**:
For the compound XY₂:
- Given: \(\Delta T_f = 2.3^\circ C\) and mass of solute = 1 g
- Using the formula:
\[ 2.3 = 1 \cdot 5.1 \cdot m_1 \]
Rearranging gives:
\[ m_1 = \frac{2.3}{5.1} = 0.45098 \text{ mol/kg} \]
Since the mass of solvent (benzene) is 20 g or 0.02 kg, we can find the number of moles of solute:
\[ m_1 = \frac{\text{moles of solute}}{0.02} \implies \text{moles of solute} = m_1 \cdot 0.02 = 0.45098 \cdot 0.02 = 0.0090196 \text{ moles} \]
The molar mass \(M_1\) of XY₂ can be calculated as:
\[ M_1 = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1 \text{ g}}{0.0090196 \text{ moles}} \approx 110.87 \text{ g/mol} \]

3. **Calculating Molality for XY₄**:
For the compound XY₄:
- Given: \(\Delta T_f = 1.3^\circ C\) and mass of solute = 1 g
- Using the formula:
\[ 1.3 = 1 \cdot 5.1 \cdot m_2 \]
Rearranging gives:
\[ m_2 = \frac{1.3}{5.1} = 0.25490 \text{ mol/kg} \]
The number of moles of solute is:
\[ m_2 = \frac{\text{moles of solute}}{0.02} \implies \text{moles of solute} = m_2 \cdot 0.02 = 0.25490 \cdot 0.02 = 0.005098 \text{ moles} \]
The molar mass \(M_2\) of XY₄ can be calculated as:
\[ M_2 = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1 \text{ g}}{0.005098 \text{ moles}} \approx 196.15 \text{ g/mol} \]

4. **Setting Up the Equations**:
Now we have two equations based on the molar masses:
- For XY₂: \(x + 2y = 110.87\) (Equation 1)
- For XY₄: \(x + 4y = 196.15\) (Equation 2)

5. **Solving the Equations**:
Subtract Equation 1 from Equation 2:
\[ (x + 4y) - (x + 2y) = 196.15 - 110.87 \]
This simplifies to:
\[ 2y = 85.28 \implies y = \frac{85.28}{2} = 42.64 \]

6. **Finding x**:
Substitute \(y\) back into Equation 1:
\[ x + 2(42.64) = 110.87 \]
\[ x + 85.28 = 110.87 \implies x = 110.87 - 85.28 = 25.59 \]

### Final Result:
The atomic masses of elements X and Y are:
- Atomic mass of X = 25.59 g/mol
- Atomic mass of Y = 42.64 g/mol