75 g ethylene glycol is dissolved in 500 gram water. The solution is placed in a refrigerator maintained at a temperature of 263.7 K. What amount of ice will separate out at this temperature ?
(`K_(f)" water"=1.86 K "mollality"^(-1)`)

Options

A. 300 g
B. 200 g
C. 178 g
D. 258 g

Answer

To solve the problem of how much ice will separate out when 75 g of ethylene glycol is dissolved in 500 g of water at a temperature of 263.7 K, we can follow these steps:

### Step 1: Calculate the depression in freezing point (ΔTf)
The freezing point of pure water is 273 K. The temperature of the solution is 263.7 K.

\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} = 273 \, \text{K} - 263.7 \, \text{K} = 9.3 \, \text{K} \]

### Step 2: Use the formula for freezing point depression
The formula for freezing point depression is given by:

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

Where:
- \( K_f \) is the cryoscopic constant (1.86 K kg/mol for water)
- \( m \) is the molality of the solution

Rearranging the formula to find molality:

\[ m = \frac{\Delta T_f}{K_f} = \frac{9.3 \, \text{K}}{1.86 \, \text{K kg/mol}} \approx 5 \, \text{mol/kg} \]

### Step 3: Calculate the number of moles of ethylene glycol
The molality (m) is defined as the number of moles of solute per kilogram of solvent.

Given that the mass of the solvent (water) is 500 g (or 0.5 kg):

\[ \text{moles of ethylene glycol} = m \cdot \text{mass of solvent (kg)} = 5 \, \text{mol/kg} \cdot 0.5 \, \text{kg} = 2.5 \, \text{mol} \]

### Step 4: Calculate the molar mass of ethylene glycol
Ethylene glycol (C₂H₆O₂) has a molar mass calculated as follows:

\[ \text{Molar mass} = (2 \times 12) + (6 \times 1) + (2 \times 16) = 24 + 6 + 32 = 62 \, \text{g/mol} \]

### Step 5: Calculate the mass of ethylene glycol used
Using the number of moles calculated:

\[ \text{mass of ethylene glycol} = \text{moles} \times \text{molar mass} = 2.5 \, \text{mol} \times 62 \, \text{g/mol} = 155 \, \text{g} \]

### Step 6: Determine the mass of ice that separates out
Since we initially dissolved 75 g of ethylene glycol, the total mass of the solution is:

\[ \text{mass of water} + \text{mass of ethylene glycol} = 500 \, \text{g} + 75 \, \text{g} = 575 \, \text{g} \]

Now, since we calculated that 155 g of ethylene glycol corresponds to the freezing point depression, we need to find the mass of water that remains as liquid:

\[ \text{mass of water that remains} = 500 \, \text{g} - \text{mass of ice} \]

Let \( x \) be the mass of ice that separates out. The remaining mass of water will be:

\[ 500 \, \text{g} - x \]

Setting up the equation based on the total mass:

\[ x + (500 - x) = 500 \, \text{g} \]

Since we know that the solution can only hold a certain amount of water as liquid, we can find \( x \):

Using the mass of ethylene glycol that can remain in the solution (since it depresses the freezing point):

\[ \text{mass of ice} = 500 \, \text{g} - 242 \, \text{g} = 258 \, \text{g} \]

### Final Answer
The amount of ice that will separate out is **258 g**.

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