The molal freezing point constant of water is `1.86 K m^(-1)`. If `342 g` of cane sugar `(C_(12)H_(22)O_(11))` is dissolved in `1000 g` of water, the solution will freeze at

Options

A. `-1.86^(@)C`
B. `1.86^(@)`
C. `-3.92^(@)C`
D. `2.42^(@)C`

Explanation

To find the freezing point of the solution when 342 g of cane sugar (C₁₂H₂₂O₁₁) is dissolved in 1000 g of water, we will follow these steps:

### Step 1: Identify the given data
- Mass of solute (cane sugar) = 342 g
- Molar mass of cane sugar (C₁₂H₂₂O₁₁) = 342 g/mol
- Mass of solvent (water) = 1000 g = 1 kg
- Freezing point constant of water (Kf) = 1.86 K kg⁻¹

### Step 2: Calculate the number of moles of solute
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass of solute}}{\text{molar mass of solute}} \]
\[ \text{Number of moles} = \frac{342 \, \text{g}}{342 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 3: Calculate the molality (M)
Molality (M) is defined as the number of moles of solute per kilogram of solvent:
\[ M = \frac{\text{Number of moles of solute}}{\text{mass of solvent in kg}} = \frac{1 \, \text{mol}}{1 \, \text{kg}} = 1 \, \text{mol/kg} \]

### Step 4: Calculate the change in freezing point (ΔTf)
Using the formula:
\[ \Delta T_f = K_f \times M \]
\[ \Delta T_f = 1.86 \, \text{K kg}^{-1} \times 1 \, \text{mol/kg} = 1.86 \, \text{K} \]

### Step 5: Calculate the freezing point of the solution
The freezing point of the solution (Tf) can be calculated using:
\[ T_f = T_{f0} - \Delta T_f \]
Where \(T_{f0}\) is the freezing point of pure water (0°C):
\[ T_f = 0°C - 1.86°C = -1.86°C \]

### Final Answer
The solution will freeze at **-1.86°C**.

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