It has been found that minimum temperature recorded on a hill station is `-10^(@)C`. The amount of glycerine that must be added to `40dm^(3)` water (used in car radiator) so that it does not freeze is: `[DeltaH_("fusion") = 6.01 kJ mol^(-1)]`

Options

A. 39.78 kg
B. 22.45 kg
C. 19.89 kg
D. 42.66 kg

Answer

To solve the problem, we need to determine the amount of glycerine (C3H8O3) that must be added to 40 dm³ of water to prevent it from freezing at -10°C. We will use the freezing point depression formula and the given data.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Minimum temperature (T_min) = -10°C
- Volume of water (V) = 40 dm³ = 40,000 g (since the density of water is approximately 1 g/cm³)
- ΔH_fusion (enthalpy of fusion for glycerine) = 6.01 kJ/mol
- Freezing point of pure water (T_f) = 0°C

2. **Calculate the Freezing Point Depression (ΔTf):**
\[ \Delta T_f = T_f - T_{min} = 0 - (-10) = 10°C \]

3. **Use the Freezing Point Depression Formula:**
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( i \) = van 't Hoff factor (for glycerine, \( i = 1 \))
- \( K_f \) = freezing point depression constant for water (approximately 1.86 °C kg/mol)
- \( m \) = molality of the solution

4. **Rearranging the Formula to Find Molality (m):**
\[ m = \frac{\Delta T_f}{i \cdot K_f} = \frac{10}{1 \cdot 1.86} \approx 5.38 \text{ mol/kg} \]

5. **Calculate the Number of Moles of Glycerine Required:**
Since molality (m) is defined as moles of solute per kg of solvent, we can find the moles of glycerine needed for 40 kg of water:
\[ \text{Moles of glycerine} = m \cdot \text{mass of solvent (kg)} = 5.38 \cdot 40 \approx 215.2 \text{ moles} \]

6. **Calculate the Mass of Glycerine Required:**
The molar mass of glycerine (C3H8O3) is:
\[ \text{Molar mass} = (3 \cdot 12.01) + (8 \cdot 1.008) + (3 \cdot 16.00) \approx 92.09 \text{ g/mol} \]
Now, calculate the mass of glycerine needed:
\[ \text{Mass of glycerine} = \text{Moles} \cdot \text{Molar mass} = 215.2 \cdot 92.09 \approx 19849.4 \text{ g} \approx 19.85 \text{ kg} \]

### Final Answer:
The amount of glycerine that must be added to 40 dm³ of water so that it does not freeze is approximately **19.85 kg**.