How many grams of methyl alcohol should be added to 10 litre tank of water to prevent its freezing at `268K`?
`(K_(f)` for water is `1.86 K kg mol^(-1))`

Options

A. `899.04g`
B. `886.02g`
C. `868.06g`
D. `880.07g`

Explanation

To solve the problem of how many grams of methyl alcohol should be added to a 10-liter tank of water to prevent it from freezing at 268 K, we will follow these steps:

### Step 1: Gather Given Data
- Freezing point of pure water (Tf) = 273 K (0 °C)
- Freezing point of the solution (Tf solution) = 268 K
- Kf (cryoscopic constant for water) = 1.86 K kg mol⁻¹
- Volume of water = 10 L
- Density of water = 1 g/mL → Mass of water = 10 L × 1000 g/L = 10,000 g = 10 kg
- Molecular weight of methyl alcohol (CH₃OH) = 32 g/mol

### Step 2: Calculate the Change in Freezing Point (ΔTf)
\[ \Delta T_f = T_f (\text{pure solvent}) - T_f (\text{solution}) = 273 \, \text{K} - 268 \, \text{K} = 5 \, \text{K} \]

### Step 3: Use the Freezing Point Depression Formula
The formula for freezing point depression is given by:
\[ \Delta T_f = K_f \times m \]
Where:
- \( m \) is the molality of the solution.

### Step 4: Calculate Molality (m)
Rearranging the formula to find molality:
\[ m = \frac{\Delta T_f}{K_f} = \frac{5 \, \text{K}}{1.86 \, \text{K kg mol}^{-1}} \approx 2.688 \, \text{mol/kg} \]

### Step 5: Calculate Moles of Methyl Alcohol Required
Using the definition of molality:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Let the number of moles of methyl alcohol be \( n \):
\[ 2.688 = \frac{n}{10 \, \text{kg}} \implies n = 2.688 \times 10 = 26.88 \, \text{mol} \]

### Step 6: Calculate Mass of Methyl Alcohol Required
Using the molecular weight of methyl alcohol:
\[ \text{mass of methyl alcohol} = n \times \text{molecular weight} = 26.88 \, \text{mol} \times 32 \, \text{g/mol} \approx 860.16 \, \text{g} \]

### Step 7: Final Answer
Thus, the mass of methyl alcohol that should be added is approximately **860.16 grams**.