`A` solution contains `68.4gms` of cane sugar `(C_(12)H_(22)O_(11))` in `1000gms` of water. Calculate the following for this solution (a) Vapour pressure: (b) Osmotic pressure at `20^(@)C`, (c) Freezing point, (d) Boiling point. [density of the solution `=1.024gm cm^(-3)`, vapour pressure of water `=17.54mm`, latent heat of fusion `=80 "cal" cm^(-1)`]`

Explanation

To solve the question step by step, we will calculate the vapor pressure, osmotic pressure, freezing point, and boiling point of the cane sugar solution.

### Given Data:
- Mass of cane sugar (C₁₂H₂₂O₁₁) = 68.4 g
- Mass of water = 1000 g
- Density of solution = 1.024 g/cm³
- Vapour pressure of pure water (P₀) = 17.54 mmHg
- Latent heat of fusion (L_f) = 80 cal/g

### Step 1: Calculate the number of moles of solute (cane sugar)
1. **Molar mass of cane sugar (C₁₂H₂₂O₁₁)**:
- C: 12 × 12.01 g/mol = 144.12 g/mol
- H: 22 × 1.008 g/mol = 22.176 g/mol
- O: 11 × 16.00 g/mol = 176.00 g/mol
- Total = 144.12 + 22.176 + 176.00 = 342.296 g/mol (approximately 342 g/mol)

2. **Number of moles of cane sugar (n_b)**:
\[ n_b = \frac{\text{mass of cane sugar}}{\text{molar mass}} = \frac{68.4 \, \text{g}}{342 \, \text{g/mol}} \approx 0.2 \, \text{mol} \]

### Step 2: Calculate the number of moles of solvent (water)
1. **Molar mass of water (H₂O)** = 18 g/mol

2. **Number of moles of water (n_a)**:
\[ n_a = \frac{\text{mass of water}}{\text{molar mass}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 3: Calculate the mole fraction of the solvent
1. **Mole fraction of water (X_a)**:
\[ X_a = \frac{n_a}{n_a + n_b} = \frac{55.56}{55.56 + 0.2} \approx \frac{55.56}{55.76} \approx 0.995 \]

### Step 4: Calculate the vapor pressure of the solution (P_a)
1. **Using Raoult's Law**:
\[ P_a = P_0 \cdot X_a = 17.54 \, \text{mmHg} \cdot 0.995 \approx 17.48 \, \text{mmHg} \]

### Step 5: Calculate the osmotic pressure (π) at 20°C
1. **Calculate the volume of the solution**:
- Total mass of the solution = 1000 g + 68.4 g = 1068.4 g
- Volume of solution (V) = \(\frac{\text{mass}}{\text{density}} = \frac{1068.4 \, \text{g}}{1.024 \, \text{g/cm}^3} \approx 1043.4 \, \text{ml} = 1.0434 \, \text{L}\)

2. **Calculate molarity (C)**:
\[ C = \frac{n_b}{V} = \frac{0.2 \, \text{mol}}{1.0434 \, \text{L}} \approx 0.191 \, \text{mol/L} \]

3. **Using the formula for osmotic pressure**:
\[ \pi = C \cdot R \cdot T \]
- R = 0.0821 L·atm/(K·mol) (convert to atm)
- T = 20°C = 293 K
\[ \pi = 0.191 \, \text{mol/L} \cdot 0.0821 \, \text{L·atm/(K·mol)} \cdot 293 \, \text{K} \approx 4.611 \, \text{atm} \]

### Step 6: Calculate the freezing point depression (ΔT_f)
1. **Calculate molality (m)**:
\[ m = \frac{n_b}{\text{mass of solvent in kg}} = \frac{0.2 \, \text{mol}}{1 \, \text{kg}} = 0.2 \, \text{mol/kg} \]

2. **Calculate K_f**:
\[ K_f = \frac{R \cdot T_f^2}{1000 \cdot L_f} = \frac{2 \cdot (273)^2}{1000 \cdot 80} \approx 1.86 \, \text{°C kg/mol} \]

3. **Calculate ΔT_f**:
\[ ΔT_f = K_f \cdot m = 1.86 \cdot 0.2 \approx 0.372 \, \text{°C} \]

4. **Freezing point of the solution (T_f')**:
\[ T_f' = 0 - ΔT_f = 0 - 0.372 \approx -0.372 \, \text{°C} \]

### Step 7: Calculate the boiling point elevation (ΔT_b)
1. **Calculate K_b**:
\[ K_b = \frac{R \cdot T_b^2}{1000 \cdot L_v} = \frac{2 \cdot (373)^2}{1000 \cdot 540} \approx 0.515 \, \text{°C kg/mol} \]

2. **Calculate ΔT_b**:
\[ ΔT_b = K_b \cdot m = 0.515 \cdot 0.2 \approx 0.103 \, \text{°C} \]

3. **Boiling point of the solution (T_b')**:
\[ T_b' = 100 + ΔT_b = 100 + 0.103 \approx 101.103 \, \text{°C} \]

### Summary of Results:
- (a) Vapor pressure of the solution = 17.48 mmHg
- (b) Osmotic pressure at 20°C = 4.611 atm
- (c) Freezing point = -0.372 °C
- (d) Boiling point = 101.103 °C