How many moles of sucrose should be dissolved in `500gms` of water so as to get a solution which has a difference of `104^(@)C` between boiling point and freezing point. `(K_(f)=1.86 K Kg "mol"^(-1).K_(b)=0.52K Kg "mol"^(-1))`

Options

A. `1.08`
B. `3.36`
C. `1.80`
D. `0.840`

Explanation

To solve the problem, we need to determine how many moles of sucrose should be dissolved in 500 grams of water to achieve a solution with a boiling point and freezing point difference of 104°C. We will use the given values for the freezing point depression constant (Kf) and boiling point elevation constant (Kb).

### Step-by-Step Solution:

1. **Identify the Freezing Point Depression and Boiling Point Elevation:**
- The difference between the boiling point (Tv) and freezing point (Tf) is given as 104°C.
- The normal boiling point of water is 100°C, so we can find the freezing point:
\[ Tv - Tf = 104 \implies 100 - Tf = 104 \implies Tf = 100 - 104 = -4°C \]

2. **Calculate the Depression in Freezing Point (ΔTf):**
- The depression in freezing point (ΔTf) is calculated as:
\[ \Delta Tf = Tf_{water} - Tf = 0°C - (-4°C) = 4°C \]

3. **Use the Freezing Point Depression Formula:**
- The formula for freezing point depression is:
\[ \Delta Tf = Kf \cdot m \]
- Where:
- \( Kf = 1.86 \, \text{K kg mol}^{-1} \)
- \( m \) is the molality, which is defined as the number of moles of solute (n) per kilogram of solvent.

4. **Convert the Mass of Water to Kilograms:**
- The mass of water is given as 500 grams, which is:
\[ 500 \, \text{g} = 0.5 \, \text{kg} \]

5. **Substitute Values into the Freezing Point Depression Formula:**
- Rearranging the formula to solve for m (molality):
\[ m = \frac{\Delta Tf}{Kf} = \frac{4°C}{1.86 \, \text{K kg mol}^{-1}} \approx 2.15 \, \text{mol/kg} \]

6. **Calculate the Number of Moles of Sucrose:**
- Since molality (m) is defined as moles of solute per kilogram of solvent, we can calculate the number of moles (n):
\[ n = m \cdot \text{mass of solvent in kg} = 2.15 \, \text{mol/kg} \cdot 0.5 \, \text{kg} = 1.075 \, \text{moles} \]

7. **Final Result:**
- Rounding off, we find that approximately:
\[ n \approx 1.08 \, \text{moles} \]

### Conclusion:
The number of moles of sucrose that should be dissolved in 500 grams of water to achieve the desired boiling point and freezing point difference is approximately **1.08 moles**.