What is the freezing point of a solution contains `10.0g`of glucose`C_(6)H_(12)O_(6)`, in `100g` of `H_(2)O` ? `K_(f)=1.86^(@)C//m`

Options

A. `-0.186^(@)C`
B. `+0.186^(@)C`
C. `-0.10^(@)C`
D. `-1.03^(@)C`

Explanation

To find the freezing point of a solution containing 10.0 g of glucose (C₆H₁₂O₆) in 100 g of water, we will follow these steps:

### Step 1: Calculate the moles of glucose
1. **Determine the molar mass of glucose (C₆H₁₂O₆)**:
- Carbon (C): 12.01 g/mol × 6 = 72.06 g/mol
- Hydrogen (H): 1.008 g/mol × 12 = 12.096 g/mol
- Oxygen (O): 16.00 g/mol × 6 = 96.00 g/mol
- Total molar mass = 72.06 + 12.096 + 96.00 = 180.156 g/mol

2. **Calculate the number of moles of glucose**:
\[ \text{Moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} = \frac{10.0 \, \text{g}}{180.156 \, \text{g/mol}} \approx 0.0555 \, \text{mol} \]

### Step 2: Calculate the molality of the solution
1. **Convert the mass of water from grams to kilograms**:
\[ \text{Mass of water} = 100 \, \text{g} = 0.100 \, \text{kg} \]

2. **Calculate the molality (m)**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0555 \, \text{mol}}{0.100 \, \text{kg}} = 0.555 \, \text{mol/kg} \]

### Step 3: Calculate the depression in freezing point (ΔTf)
1. **Use the formula for depression in freezing point**:
\[ \Delta T_f = K_f \times m \]
where \( K_f = 1.86 \, ^\circ C/m \).

2. **Calculate ΔTf**:
\[ \Delta T_f = 1.86 \, ^\circ C/m \times 0.555 \, \text{mol/kg} \approx 1.033 \, ^\circ C \]

### Step 4: Calculate the freezing point of the solution
1. **The freezing point of pure water is 0 °C**. Thus, the freezing point of the solution is:
\[ \text{Freezing point of solution} = 0 \, ^\circ C - \Delta T_f = 0 \, ^\circ C - 1.033 \, ^\circ C \approx -1.033 \, ^\circ C \]

### Final Answer:
The freezing point of the solution is approximately **-1.033 °C**.

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