To solve the problem step by step, we will use the given data and apply the relevant formulas for freezing point depression and boiling point elevation.
### Step 1: Calculate the molality from the freezing point depression.
The formula for freezing point depression is given by:
\[
\Delta T_f = K_f \cdot m
\]
where:
- \(\Delta T_f\) is the depression in freezing point (0.93 K),
- \(K_f\) is the cryoscopic constant for water (1.86 K kg/mol),
- \(m\) is the molality of the solution.
Rearranging the formula to find molality:
\[
m = \frac{\Delta T_f}{K_f} = \frac{0.93}{1.86} = 0.5 \, \text{mol/kg}
\]
### Step 2: Calculate the boiling point elevation.
The formula for boiling point elevation is given by:
\[
\Delta T_b = K_b \cdot m
\]
where:
- \(\Delta T_b\) is the elevation in boiling point, which is \(100.26 - 100 = 0.26 \, \text{°C}\).
We need to find \(K_b\) for water. The relationship between \(K_f\) and \(K_b\) is:
\[
K_b = \frac{K_f \cdot 1000}{K_f + 1000}
\]
However, for simplicity, we can use the boiling point elevation directly with the molality we calculated.
### Step 3: Calculate the molality using the boiling point elevation.
Using the boiling point elevation:
\[
\Delta T_b = K_b \cdot m
\]
We can rearrange this to find \(K_b\):
\[
K_b = \frac{\Delta T_b}{m} = \frac{0.26}{0.5} = 0.52 \, \text{K kg/mol}
\]
### Step 4: Use the second boiling point elevation to find the molecular weight.
Now, we have another scenario where \(7.87 \, \text{g}\) of \(A_x\) is dissolved in \(100 \, \text{g}\) of water, and the boiling point is \(100.44 \, \text{°C}\). Thus:
\[
\Delta T_b' = 100.44 - 100 = 0.44 \, \text{°C}
\]
Using the boiling point elevation formula again:
\[
0.44 = K_b \cdot m'
\]
where \(m'\) is the new molality. Rearranging gives:
\[
m' = \frac{0.44}{0.52} = 0.846 \, \text{mol/kg}
\]
### Step 5: Calculate the molecular weight of \(A_x\).
Using the formula for molality:
\[
m' = \frac{W}{M \cdot W_{solvent}}
\]
where:
- \(W\) is the mass of the solute (7.87 g),
- \(M\) is the molar mass of \(A_x\),
- \(W_{solvent}\) is the mass of the solvent (0.1 kg for 100 g of water).
Rearranging gives:
\[
M = \frac{W}{m' \cdot W_{solvent}} = \frac{7.87}{0.846 \cdot 0.1} = \frac{7.87}{0.0846} \approx 93 \, \text{g/mol}
\]
### Step 6: Relate the molecular weight to \(x\).
The molecular weight of \(A_x\) can be expressed as:
\[
M = x \cdot (31 \, \text{g/mol}) = 93 \, \text{g/mol}
\]
Solving for \(x\):
\[
x = \frac{93}{31} = 3
\]
### Final Answer:
The value of \(x\) is \(3\).
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