The amount of ice that will separate out from a solution containing 25 g of ethylene glycol in 100 g of water that is cooled to `-10^(@)C`, will be [Given : `K_(f) "for" H_(2)O = 1.86K mol^(-1) kg`]

Options

A. 50.0 g
B. 25.0 g
C. 12.5 g
D. 30.0 g

Answer

To solve the problem of how much ice will separate out from a solution containing 25 g of ethylene glycol in 100 g of water when cooled to -10°C, we can follow these steps:

### Step 1: Calculate the molar mass of ethylene glycol (C2H6O2)
Ethylene glycol has the following molecular composition:
- Carbon (C): 2 atoms × 12 g/mol = 24 g/mol
- Hydrogen (H): 6 atoms × 1 g/mol = 6 g/mol
- Oxygen (O): 2 atoms × 16 g/mol = 32 g/mol

Adding these together gives:
\[ \text{Molar mass of ethylene glycol} = 24 + 6 + 32 = 62 \text{ g/mol} \]

### Step 2: Calculate the number of moles of ethylene glycol
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
We have:
\[ \text{Number of moles of ethylene glycol} = \frac{25 \text{ g}}{62 \text{ g/mol}} \approx 0.403 \text{ moles} \]

### Step 3: Calculate the molality of the solution
Molality (m) is defined as:
\[ \text{Molality} = \frac{\text{number of moles of solute}}{\text{mass of solvent (kg)}} \]
The mass of the solvent (water) is 100 g, which is 0.1 kg. Therefore:
\[ \text{Molality} = \frac{0.403 \text{ moles}}{0.1 \text{ kg}} = 4.03 \text{ mol/kg} \]

### Step 4: Calculate the depression in freezing point (ΔTf)
The depression in freezing point can be calculated using the formula:
\[ \Delta T_f = K_f \times m \]
Where \( K_f \) for water is given as 1.86 K kg/mol. Thus:
\[ \Delta T_f = 1.86 \text{ K kg/mol} \times 4.03 \text{ mol/kg} \approx 7.5 \text{ °C} \]

### Step 5: Determine the freezing point of the solution
The freezing point of pure water is 0°C. Therefore, the freezing point of the solution is:
\[ \text{Freezing point of solution} = 0 \text{ °C} - 7.5 \text{ °C} = -7.5 \text{ °C} \]

### Step 6: Calculate the amount of ice that will separate out
Since the solution is cooled to -10°C, which is below the freezing point of the solution (-7.5°C), ice will form. The amount of ice that separates out can be calculated by finding the difference between the initial mass of water and the mass of water that remains in the solution.

The mass of water that remains liquid at -10°C is equal to the mass of the solution at its freezing point. Since the freezing point depression is 7.5°C, we can assume that all the water will freeze out at -10°C, leading to:
\[ \text{Mass of ice formed} = \text{Initial mass of water} - \text{Mass of water in solution} \]

Assuming that the entire solution freezes, we can calculate:
\[ \text{Mass of ice} = 100 \text{ g} - (100 \text{ g} - 25 \text{ g}) = 25 \text{ g} \]

### Final Answer
The amount of ice that will separate out is **25 g**.

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