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 solve the problem of determining the freezing point of a solution when 342 g of cane sugar (C₁₂H₂₂O₁₁) is dissolved in 1000 g of water, we can follow these steps:

### Step 1: Determine the molar mass of cane sugar
The molar mass of cane sugar (C₁₂H₂₂O₁₁) is given as 342 g/mol.

### Step 2: Calculate the number of moles of cane sugar
To find the number of moles of cane sugar, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{342 \, \text{g}}{342 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We have 1000 g of water, which is 1 kg. Therefore:
\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{mass of solvent (kg)}} = \frac{1 \, \text{mol}}{1 \, \text{kg}} = 1 \, \text{m} \]

### Step 4: Calculate the depression in freezing point (ΔTf)
The formula for the depression in freezing point is given by:
\[ \Delta T_f = K_f \times m \]
Where \( K_f \) is the molal freezing point constant of water (1.86 K kg/mol). Substituting the values:
\[ \Delta T_f = 1.86 \, \text{K kg/mol} \times 1 \, \text{m} = 1.86 \, \text{K} \]

### Step 5: Determine the freezing point of the solution
The freezing point of pure water is 0 °C. The freezing point of the solution can be calculated as:
\[ T_f = T_f^0 - \Delta T_f \]
Substituting the values:
\[ T_f = 0 \, \text{°C} - 1.86 \, \text{°C} = -1.86 \, \text{°C} \]

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

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