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Step-by-Step Solution
Step 1: Understand the Given Data
• Mass of solute (ethylene glycol), msolute = 62 g
• Mass of water (solvent), msolvent = 250 g
• Freezing point depression constant for water, Kf = 1.86 K kg mol−1
• Final temperature to which the solution is cooled, T = −10 °C
• Molar mass of ethylene glycol, Meg = 62 g mol−1 (since 62 g corresponds to 1 mole)
Step 2: Calculate the Freezing Point of the Original Solution
The depression in freezing point ΔTf for an ideal solution is given by:
$\Delta T_f = K_f \times m$
where m is the molality (moles of solute per kg of solvent).
First, calculate ΔTf for the entire initial mixture (62 g ethylene glycol + 250 g water):
Number of moles of ethylene glycol = $n_{\text{solute}} = \frac{62\text{ g}}{62\text{ g mol}^{-1}} = 1\text{ mol}$
Mass of solvent in kg = $\frac{250\text{ g}}{1000} = 0.250\text{ kg}$
Molality, $m = \frac{n_{\text{solute}}}{\text{kg of solvent}} = \frac{1}{0.250} = 4\text{ mol kg}^{-1}$
Thus,
$$
\Delta T_f = K_f \times m = 1.86 \times 4 = 7.44\text{ K}
$$
The freezing point of pure water is 0 °C. Therefore, the freezing point Tf of the original solution is:
$$
T_f = 0^\circ\text{C} - 7.44^\circ\text{C} = -7.44^\circ\text{C}
$$
Step 3: Recognize the Further Cooling and Ice Formation
The solution will not start to freeze until it reaches its freezing point of −7.44 °C. Now the temperature is lowered further to −10 °C. Below −7.44 °C, some of the water will separate out as ice, so the liquid portion that remains in solution will have less solvent (water). This in turn increases the molality of the remaining liquid solution.
Step 4: Determine the Mass of Water Remaining in Liquid Phase
Let Wl be the mass of water (in g) still present in the liquid solution at −10 °C. The new freezing point depression ΔTf,new is 10 °C because the solution’s freezing point is at −10 °C now:
$$
\Delta T_{f,\text{new}} = 10 = K_f \times \frac{n_{\text{solute}}}{\text{kg of solvent in liquid}}
$$
Since the solute (ethylene glycol) does not freeze out, its moles remain the same (1 mole). Hence,
$$
10 = 1.86 \times \frac{1}{W_l/1000}
$$
Here, $W_l/1000$ is the mass of liquid water in kilograms. Solving for $W_l$:
$$
10 = 1.86 \times \frac{1000}{W_l} \quad \Longrightarrow \quad
\frac{10 \times W_l}{1.86} = 1000 \quad \Longrightarrow \quad
W_l = \frac{1000 \times 1.86}{10} = 186\text{ g}
$$
Therefore, 186 g of water remains in the liquid phase at −10 °C.
Step 5: Calculate the Amount of Water Separated as Ice
Initially, there were 250 g of water. If 186 g of water remain in the liquid phase, the rest has frozen out as ice:
$$
\text{Mass of water frozen as ice} = 250\text{ g} - 186\text{ g} = 64\text{ g}
$$
Hence, 64 g of water separate out as ice.
Final Answer
64 g of water is separated as ice.
