The amount (in grams) of sucrose (mol.wt. = 342g) that should be dissolved in 100 g water in order to produce a solution with a `105.0^@C` difference between the boiling point and freezing point is (Given that `k_f=1.86Kkgmol^(-1) and k_b=0.52Kkgmol^(-1)" for water")` Report your answer by rounding it up to to the nearest whole number.

Answer

To solve the problem, we need to determine the amount of sucrose that should be dissolved in 100 g of water to achieve a boiling point elevation and freezing point depression that results in a total temperature difference of 105.0 °C.

### Step-by-Step Solution:

1. **Identify the Constants and Given Values:**
- Molecular weight of sucrose (Mw) = 342 g/mol
- Mass of water (solvent) = 100 g = 0.1 kg
- Boiling point elevation constant (Kb) = 0.52 K kg/mol
- Freezing point depression constant (Kf) = 1.86 K kg/mol
- Total temperature difference (ΔT) = 105.0 °C

2. **Set Up the Equations:**
- The boiling point elevation (ΔTb) and freezing point depression (ΔTf) can be expressed as:
\[ \Delta T_b = K_b \cdot m \]
\[ \Delta T_f = K_f \cdot m \]
- The total temperature difference is given by:
\[ \Delta T_b + |\Delta T_f| = 105.0 \]

3. **Express the Total Temperature Difference:**
- Since ΔTf is a depression, it is negative, so we can write:
\[ K_b \cdot m + K_f \cdot m = 105.0 \]
- This simplifies to:
\[ m(K_b + K_f) = 105.0 \]

4. **Calculate the Molality (m):**
- Substitute the values of Kb and Kf:
\[ m(0.52 + 1.86) = 105.0 \]
\[ m(2.38) = 105.0 \]
\[ m = \frac{105.0}{2.38} \approx 44.12 \text{ mol/kg} \]

5. **Relate Molality to Moles of Solute:**
- Molality (m) is defined as moles of solute per kg of solvent:
\[ m = \frac{\text{moles of sucrose}}{0.1 \text{ kg}} \]
- Therefore, the moles of sucrose can be calculated as:
\[ \text{moles of sucrose} = m \times 0.1 \approx 44.12 \times 0.1 = 4.412 \text{ moles} \]

6. **Convert Moles of Sucrose to Grams:**
- Using the molecular weight of sucrose:
\[ \text{mass of sucrose} = \text{moles of sucrose} \times \text{molecular weight} \]
\[ \text{mass of sucrose} = 4.412 \times 342 \approx 1508.544 \text{ grams} \]

7. **Round to the Nearest Whole Number:**
- The final answer is:
\[ \text{mass of sucrose} \approx 1509 \text{ grams} \]

### Final Answer:
The amount of sucrose that should be dissolved in 100 g of water is approximately **1509 grams**.