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 an aqueous solution containing 17 g of \( C_2H_5OH \) (ethanol) in 100 g of water, we will follow these steps:

### Step 1: Calculate the number of moles of \( C_2H_5OH \)

The formula to calculate the number of moles is:

\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]

The molar mass of \( C_2H_5OH \) (ethanol) is calculated as follows:

- Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
- Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol
- Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol

Adding these together gives:

\[ \text{Molar mass of } C_2H_5OH = 24.02 + 6.048 + 16.00 = 46.068 \approx 46 \text{ g/mol} \]

Now, substituting the values:

\[ \text{Number of moles} = \frac{17 \text{ g}}{46 \text{ g/mol}} \approx 0.3696 \text{ moles} \]

### Step 2: Calculate the mass of the solvent in kg

The mass of the solvent (water) is given as 100 g. To convert this to kg:

\[ \text{Mass of solvent (kg)} = \frac{100 \text{ g}}{1000} = 0.1 \text{ kg} \]

### Step 3: Calculate the molality of the solution

Molality (m) is defined as the number of moles of solute per kg of solvent:

\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent (kg)}} \]

Substituting the values:

\[ \text{Molality} = \frac{0.3696 \text{ moles}}{0.1 \text{ kg}} = 3.696 \text{ mol/kg} \]

### Step 4: Calculate the depression in freezing point (\( \Delta T_f \))

The depression in freezing point is calculated using the formula:

\[ \Delta T_f = K_f \times m \]

Where \( K_f \) for water is given as 1.86 K kg/mol. Therefore:

\[ \Delta T_f = 1.86 \text{ K kg/mol} \times 3.696 \text{ mol/kg} \approx 6.87 \text{ K} \]

### Step 5: Calculate the new freezing point

The freezing point of pure water is 0 °C. The new freezing point (\( T_f \)) can be calculated as:

\[ T_f = T_i - \Delta T_f \]

Where \( T_i \) is the initial freezing point (0 °C):

\[ T_f = 0 °C - 6.87 °C \approx -6.87 °C \]

### Final Answer

The freezing point of the aqueous solution containing 17 g of \( C_2H_5OH \) in 100 g of water is approximately **-6.87 °C**.

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