The molar freezing point constant for water is `1.86 ^@C`/ molal . If 342 gm of canesugar `(C_(12)H_(22)O_(11))` are dissolved in 1000 gm of water, the solution will freeze at

Options

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

Answer

To solve the problem, we need to calculate the freezing point depression when 342 g of cane sugar (C₁₂H₂₂O₁₁) is dissolved in 1000 g of water. We will use the formula for freezing point depression:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of cane sugar (C₁₂H₂₂O₁₁) = 342 g
- Mass of water (solvent) = 1000 g
- Molar freezing point constant for water (Kf) = 1.86 °C/molal

2. **Calculate the Molar Mass of Cane Sugar (C₁₂H₂₂O₁₁):**
- Carbon (C): 12 × 12.01 g/mol = 144.12 g/mol
- Hydrogen (H): 22 × 1.008 g/mol = 22.176 g/mol
- Oxygen (O): 11 × 16.00 g/mol = 176.00 g/mol
- Total molar mass = 144.12 + 22.176 + 176.00 = 342.296 g/mol (approximately 342 g/mol)

3. **Calculate the Number of Moles of Cane Sugar:**
\[ \text{Number of moles (n)} = \frac{\text{mass of solute}}{\text{molar mass}} = \frac{342 \text{ g}}{342 \text{ g/mol}} = 1 \text{ mol} \]

4. **Convert the Mass of Solvent to Kilograms:**
\[ \text{Mass of water in kg} = \frac{1000 \text{ g}}{1000} = 1 \text{ kg} \]

5. **Calculate the Molality (M):**
\[ \text{Molality (M)} = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} = \frac{1 \text{ mol}}{1 \text{ kg}} = 1 \text{ molal} \]

6. **Calculate the Freezing Point Depression (ΔTf):**
\[ \Delta Tf = Kf \times M = 1.86 \text{ °C/molal} \times 1 \text{ molal} = 1.86 \text{ °C} \]

7. **Determine the Freezing Point of the Solution:**
- The normal freezing point of water is 0 °C.
- Therefore, the freezing point of the solution will be:
\[ \text{Freezing point} = 0 \text{ °C} - \Delta Tf = 0 \text{ °C} - 1.86 \text{ °C} = -1.86 \text{ °C} \]

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

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