0.0125 mol of sucrose is dissolved in 100 gm of water and it undergo partial inversion according to following equation
`C_12H_22O_11+H_2OtoC_6H_12O_6+C_6H_12O_6`
If elevation in boiling point of solution is `0.104^@C` calculate `1/10` mol percentage of sugar inverted `(K_b,_(H_2O)=0.52)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 + C_6H_{12}O_6 \]
This means that one mole of sucrose (C₁₂H₂₂O₁₁) reacts with water to produce two moles of glucose (C₆H₁₂O₆).

### Step 2: Calculate the Molality
We know that molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

Here, we have:
- Moles of sucrose = 0.0125 mol
- Mass of water = 100 g = 0.1 kg

Thus, the molality of the solution is:
\[ m = \frac{0.0125 \text{ mol}}{0.1 \text{ kg}} = 0.125 \text{ mol/kg} \]

### Step 3: Use the Elevation in Boiling Point Formula
The elevation in boiling point (\(\Delta T_b\)) is given by:
\[ \Delta T_b = K_b \cdot m \]
Where:
- \(K_b\) for water = 0.52 °C kg/mol
- \(\Delta T_b\) = 0.104 °C

Substituting the values:
\[ 0.104 = 0.52 \cdot m \]

### Step 4: Calculate the Total Molality
Rearranging the formula to find molality:
\[ m = \frac{0.104}{0.52} = 0.2 \text{ mol/kg} \]

### Step 5: Set Up the Equation for Moles
Let \(x\) be the number of moles of sucrose that undergo inversion. After inversion:
- Moles of sucrose remaining = \(0.0125 - x\)
- Moles of glucose produced = \(2x\) (since each mole of sucrose produces 2 moles of glucose)

The total moles in the solution after inversion is:
\[ (0.0125 - x) + 2x = 0.0125 + x \]

### Step 6: Relate to Molality
Now we can express the molality in terms of \(x\):
\[ \text{Molality} = \frac{0.0125 + x}{0.1} \]

Setting this equal to the previously calculated molality:
\[ \frac{0.0125 + x}{0.1} = 0.2 \]

### Step 7: Solve for \(x\)
Multiplying both sides by 0.1:
\[ 0.0125 + x = 0.02 \]
\[ x = 0.02 - 0.0125 = 0.0075 \text{ mol} \]

### Step 8: Calculate the Percentage of Sugar Inverted
The percentage of sucrose inverted is given by:
\[ \text{Percentage inverted} = \left(\frac{x}{\text{initial moles of sucrose}}\right) \times 100 \]
\[ \text{Percentage inverted} = \left(\frac{0.0075}{0.0125}\right) \times 100 = 60\% \]

### Step 9: Calculate \( \frac{1}{10} \) of the Mole Percentage
Finally, we need to find \( \frac{1}{10} \) of the mole percentage:
\[ \frac{1}{10} \times 60\% = 6\% \]

Thus, the final answer is:
**6%**