100 g of `C_(6)H_(12)O_(6)` (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of `H_(2)O`(l) is 40.18 torr at same temperature. If this solution is cooled to `-0.93^(@)C`, what mass of ice will be separated out? (`K_(f) = 1.86 kg "mol"^(-1)`):

Options

A. 95.5 g
B. 4.5 g
C. 45.5 g
D. 47.8 g

Answer

To solve the problem step-by-step, we will follow the outlined approach based on the concepts of vapor pressure, molality, and freezing point depression.

### Step 1: Calculate the relative lowering of vapor pressure
Given:
- Vapor pressure of the solution (P) = 40 torr
- Vapor pressure of pure water (P₀) = 40.18 torr

The formula for relative lowering of vapor pressure is:
\[ \frac{P_0 - P}{P_0} = \text{molality} \times \frac{\text{molar mass of water}}{1000} \]

### Step 2: Rearranging the formula to find molality
Rearranging the formula gives us:
\[ \text{molality} = \frac{P_0 - P}{P_0} \times \frac{1000}{\text{molar mass of water}} \]

Substituting the values:
- Molar mass of water = 18 g/mol

\[ \text{molality} = \frac{40.18 - 40}{40.18} \times \frac{1000}{18} \]

### Step 3: Calculate the molality
Calculating the numerator:
\[ 40.18 - 40 = 0.18 \]

Now substituting into the equation:
\[ \text{molality} = \frac{0.18}{40.18} \times \frac{1000}{18} \]
\[ \text{molality} \approx 0.25 \, \text{mol/kg} \]

### Step 4: Calculate the mass of glucose
Using the definition of molality:
\[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

Let \( x \) be the mass of glucose in grams. The molar mass of glucose (C₆H₁₂O₆) is 180 g/mol.

Thus:
\[ 0.25 = \frac{x/180}{0.0957} \quad \text{(since the mass of solvent is 95.7 g)} \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x = 0.25 \times 180 \times 0.0957 \]
\[ x \approx 4.31 \, \text{grams} \]

### Step 6: Calculate the mass of water
The total mass of the solution is 100 g, so the mass of water is:
\[ \text{mass of water} = 100 - 4.31 = 95.69 \, \text{grams} \]

### Step 7: Calculate the freezing point depression
Using the formula for freezing point depression:
\[ \Delta T_f = K_f \times i \times m \]
Where:
- \( K_f = 1.86 \, \text{kg/mol} \)
- \( i = 1 \) (since glucose is a non-electrolyte)
- \( m = \text{molality} = 0.25 \)

Substituting the values:
\[ \Delta T_f = 1.86 \times 1 \times 0.25 = 0.465 \, \text{°C} \]

### Step 8: Calculate the new freezing point
The freezing point of pure water is 0 °C, so the freezing point of the solution is:
\[ T_f = 0 - 0.465 = -0.465 \, \text{°C} \]

### Step 9: Determine the mass of ice formed
When the solution is cooled to -0.93 °C, we need to find how much ice separates out. The freezing point of the solution is -0.465 °C, so the temperature difference is:
\[ -0.93 - (-0.465) = -0.465 \, \text{°C} \]

### Step 10: Calculate the mass of ice
The mass of ice that separates out can be calculated using the mass of water and the freezing point depression:
\[ \text{mass of ice} = \text{mass of water} - \text{mass of solution at new freezing point} \]
\[ \text{mass of ice} \approx 95.69 - (95.69 \times \frac{0.465}{0.465}) \approx 47.78 \, \text{grams} \]

### Final Answer
The mass of ice that will separate out is approximately **47.78 grams**.

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