To solve the problem, we will follow these steps:
### Step 1: Understand the Depression in Freezing Point Formula
The depression in freezing point (ΔTf) is given by the formula:
\[
\Delta T_f = i \cdot K_f \cdot m
\]
where:
- \(i\) = van 't Hoff factor (for non-electrolytes, \(i = 1\))
- \(K_f\) = molal depression constant (given as 5.1 for benzene)
- \(m\) = molality of the solution
### Step 2: Calculate the Molar Mass of \(XY_2\)
For the compound \(XY_2\):
- Given: 1 g of \(XY_2\) lowers the freezing point by \(2.3^\circ C\).
- We can rearrange the formula to find molality:
\[
m = \frac{\Delta T_f}{K_f} = \frac{2.3}{5.1}
\]
Calculating this gives:
\[
m = 0.45098 \, \text{mol/kg}
\]
Next, we calculate the molality using the definition:
\[
m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{n}{0.020}
\]
where \(0.020 \, \text{kg}\) is the mass of benzene.
Thus, we have:
\[
0.45098 = \frac{n}{0.020} \implies n = 0.45098 \times 0.020 = 0.0090196 \, \text{mol}
\]
Now, using the number of moles to find the molar mass \(M_1\) of \(XY_2\):
\[
M_1 = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1 \, \text{g}}{0.0090196 \, \text{mol}} \approx 110.87 \, \text{g/mol}
\]
### Step 3: Calculate the Molar Mass of \(XY_4\)
For the compound \(XY_4\):
- Given: 1 g of \(XY_4\) lowers the freezing point by \(1.3^\circ C\).
- Using the same formula:
\[
m = \frac{1.3}{5.1} = 0.25490 \, \text{mol/kg}
\]
Calculating the number of moles:
\[
0.25490 = \frac{n}{0.020} \implies n = 0.25490 \times 0.020 = 0.005098 \, \text{mol}
\]
Now, using the number of moles to find the molar mass \(M_2\) of \(XY_4\):
\[
M_2 = \frac{1 \, \text{g}}{0.005098 \, \text{mol}} \approx 196.15 \, \text{g/mol}
\]
### Step 4: Set Up the Equations for Atomic Masses
From the molar masses we have:
1. For \(XY_2\): \(M_1 = M_X + 2M_Y = 110.87\)
2. For \(XY_4\): \(M_2 = M_X + 4M_Y = 196.15\)
### Step 5: Solve the System of Equations
Subtract the first equation from the second:
\[
(M_X + 4M_Y) - (M_X + 2M_Y) = 196.15 - 110.87
\]
This simplifies to:
\[
2M_Y = 85.28 \implies M_Y = \frac{85.28}{2} = 42.64 \, \text{g/mol}
\]
Now substitute \(M_Y\) back into the first equation:
\[
M_X + 2(42.64) = 110.87 \implies M_X + 85.28 = 110.87 \implies M_X = 110.87 - 85.28 = 25.59 \, \text{g/mol}
\]
### Final Result
The atomic masses of \(X\) and \(Y\) are:
- \(M_X \approx 25.59 \, \text{g/mol}\)
- \(M_Y \approx 42.64 \, \text{g/mol}\)
