What would be the freezing point of aqueous solution containing 17 g of `C_(2)H_(5)OH` in 100 g of water `(K_(f) H_(2)O = 1.86 K mol^(-1)kg)` :

Options

A. `-6.8^(@)C`
B. `-0.34^(@)C`
C. `0.0^(@)C`
D. `0.34^(@)C`

Answer

To find the freezing point of the aqueous solution containing 17 g of \( C_2H_5OH \) (ethanol) in 100 g of water, we will follow these steps:

### Step 1: Calculate the Molar Mass of Ethanol (\( C_2H_5OH \))
The molecular formula for ethanol is \( C_2H_5OH \). To calculate its molar mass:
- Carbon (C): 2 atoms × 12 g/mol = 24 g/mol
- Hydrogen (H): 6 atoms × 1 g/mol = 6 g/mol
- Oxygen (O): 1 atom × 16 g/mol = 16 g/mol

Adding these together:
\[ \text{Molar mass of } C_2H_5OH = 24 + 6 + 16 = 46 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of Ethanol
Using the formula for moles:
\[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]
Substituting the values:
\[ \text{Moles of } C_2H_5OH = \frac{17 \text{ g}}{46 \text{ g/mol}} \approx 0.369 \text{ moles} \]

### Step 3: Convert the Mass of Water to Kilograms
Since the freezing point depression constant (\( K_f \)) is given in kg, we need to convert the mass of water from grams to kilograms:
\[ \text{Mass of water} = 100 \text{ g} = \frac{100}{1000} = 0.1 \text{ kg} \]

### Step 4: Calculate the Molality of the Solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Substituting the values:
\[ \text{Molality} = \frac{0.369 \text{ moles}}{0.1 \text{ kg}} = 3.69 \text{ mol/kg} \]

### Step 5: Calculate the Depression in Freezing Point (\( \Delta T_f \))
Using the formula for freezing point depression:
\[ \Delta T_f = K_f \times \text{molality} \]
Substituting the values:
\[ \Delta T_f = 1.86 \text{ K kg}^{-1} \text{ mol}^{-1} \times 3.69 \text{ mol/kg} \approx 6.86 \text{ K} \]

### Step 6: Determine the Freezing Point of the Solution
The freezing point of pure water is 0 °C. The freezing point of the solution is given by:
\[ \text{Freezing point of solution} = \text{Freezing point of solvent} - \Delta T_f \]
Substituting the values:
\[ \text{Freezing point of solution} = 0 °C - 6.86 °C = -6.86 °C \]

Thus, the freezing point of the aqueous solution is approximately **-6.86 °C**.