Compare the weigths of methanol and glycerol, which would be required separately to lower the freezing point of one Kg water by `10^@`.

Answer

To solve the problem of comparing the weights of methanol and glycerol required to lower the freezing point of 1 kg of water by 10°C, we can follow these steps:

### Step 1: Understand the Concept
The freezing point depression (ΔTf) is directly proportional to the molality of the solution and can be expressed as:
\[ \Delta T_f = K_f \cdot m \]
where \(K_f\) is the freezing point depression constant and \(m\) is the molality of the solution.

### Step 2: Identify the Molecular Weights
- The molecular weight of methanol (CH₃OH) is approximately 32 g/mol.
- The molecular weight of glycerol (C₃H₈O₃) is approximately 92 g/mol.

### Step 3: Set Up the Relationship
Since the freezing point depression is inversely proportional to the molecular weight of the solute, we can set up the following relationship:
\[ \frac{\Delta T_f \text{ (methanol)}}{\Delta T_f \text{ (glycerol)}} = \frac{M_{\text{glycerol}}}{M_{\text{methanol}}} \]
where \(M\) represents the molecular weight.

### Step 4: Substitute the Values
Substituting the known values:
\[ \frac{10}{10} = \frac{92}{32} \]
This simplifies to:
\[ 1 = \frac{92}{32} \]

### Step 5: Calculate the Ratio
Now, we can calculate the ratio of the weights required to achieve the same freezing point depression:
\[ \frac{W_{\text{methanol}}}{W_{\text{glycerol}}} = \frac{M_{\text{glycerol}}}{M_{\text{methanol}}} = \frac{92}{32} \]
This means that for every 92 g of glycerol, we would need 32 g of methanol to achieve the same freezing point depression.

### Step 6: Conclusion
Thus, the weight of methanol required is less than that of glycerol to achieve the same lowering of the freezing point of water by 10°C.

### Final Comparison
To summarize:
- Weight of methanol required: 32 g
- Weight of glycerol required: 92 g