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 ethyl alcohol (C₂H₅OH) in 100 g of water, we will follow these steps:

### Step 1: Calculate the molar mass of ethyl alcohol (C₂H₅OH)
The molecular formula of ethyl alcohol is C₂H₅OH. We can calculate its molar mass as follows:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 6 = 6 g/mol
- Oxygen (O): 16 g/mol × 1 = 16 g/mol

Adding these together:
\[ \text{Molar mass of C₂H₅OH} = 24 + 6 + 16 = 46 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of ethyl alcohol
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
We have:
\[ \text{Number of moles of C₂H₅OH} = \frac{17 \, \text{g}}{46 \, \text{g/mol}} \approx 0.3696 \, \text{mol} \]

### Step 3: Calculate the mass of the solvent in kg
The mass of the solvent (water) is given as 100 g. We need to convert this to kg:
\[ \text{Mass of water} = \frac{100 \, \text{g}}{1000} = 0.1 \, \text{kg} \]

### Step 4: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Substituting the values:
\[ m = \frac{0.3696 \, \text{mol}}{0.1 \, \text{kg}} = 3.696 \, \text{mol/kg} \]

### Step 5: Calculate the depression in freezing point (ΔTf)
Using the formula:
\[ \Delta T_f = K_f \times m \]
Where \( K_f \) for water is given as 1.86 °C kg/mol. Thus:
\[ \Delta T_f = 1.86 \, \text{°C kg/mol} \times 3.696 \, \text{mol/kg} \approx 6.87 \, \text{°C} \]

### Step 6: Calculate 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:
\[ \text{Freezing point of solution} = \text{Freezing point of solvent} - \Delta T_f \]
Substituting the values:
\[ \text{Freezing point of solution} = 0 \, \text{°C} - 6.87 \, \text{°C} \approx -6.87 \, \text{°C} \]

### Final Answer
The freezing point of the aqueous solution is approximately -6.87 °C.

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