What is the freezing point of a solution containing `6.84g` of sucrose per `500g` of water ? `K_(f)` for water is `1.86^(@)C*m^(-1)`

Answer

To find the freezing point of a solution containing 6.84 g of sucrose in 500 g of water, we will follow these steps:

### Step 1: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. **Calculate the number of moles of sucrose:**
- The molecular weight of sucrose (C12H22O11) is approximately 342 g/mol.
- Moles of sucrose = mass of sucrose (g) / molar mass (g/mol)
\[ \text{Moles of sucrose} = \frac{6.84 \, \text{g}}{342 \, \text{g/mol}} \approx 0.0200 \, \text{mol} \]

2. **Convert the mass of water to kilograms:**
\[ \text{Mass of water} = 500 \, \text{g} = 0.500 \, \text{kg} \]

3. **Calculate molality:**
\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{0.0200 \, \text{mol}}{0.500 \, \text{kg}} = 0.0400 \, \text{mol/kg} \]

### Step 2: Calculate the depression in freezing point
The depression in freezing point (\(\Delta T_f\)) can be calculated using the formula:
\[ \Delta T_f = K_f \times m \]
Where \(K_f\) for water is given as 1.86 °C/m.

1. **Substitute the values:**
\[ \Delta T_f = 1.86 \, \text{°C/m} \times 0.0400 \, \text{mol/kg} = 0.0744 \, \text{°C} \]

### Step 3: Calculate the freezing point of the solution
The normal freezing point of pure water is 0 °C. The freezing point of the solution (\(T_f\)) can be calculated as:
\[ T_f = T_f^0 - \Delta T_f \]
Where \(T_f^0\) is the freezing point of pure water.

1. **Substitute the values:**
\[ T_f = 0 \, \text{°C} - 0.0744 \, \text{°C} = -0.0744 \, \text{°C} \]

### Final Answer:
The freezing point of the solution is approximately \(-0.0744 \, \text{°C}\).

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