This problem involves calculating the amount of water that freezes out as ice when a solution is cooled below its freezing point. The key concept is depression in freezing point, a colligative property.

Step 1: Understand Freezing Point Depression

When a non-volatile solute is added to a solvent, the freezing point of the solution is lower than that of the pure solvent. The depression in freezing point (ΔT_f) is given by:

$\mathrm{\Delta T}{}_f=K{}_{f}m$

where:

• ΔT_f = Freezing point depression = Freezing point of pure solvent - Freezing point of solution
• K_f = Molal depression constant (given as 1.86 K kg mol^-1 for water)
• m = Molality of the solution (moles of solute per kg of solvent)

Step 2: Calculate the Molality (m) of the Solution

First, find the number of moles of ethylene glycol (solute). The molar mass of ethylene glycol (C₂H₆O₂) is (2*12 + 6*1 + 2*16) = 62 g/mol.

Mass of solute = 62 g

Moles of solute = Mass / Molar mass = 62 g / 62 g/mol = 1 mole

Mass of solvent (water) = 250 g = 0.250 kg

Molality (m) = moles of solute / mass of solvent (in kg) = 1 mol / 0.250 kg = 4 mol/kg

Step 3: Calculate the Depression in Freezing Point (ΔT_f)

$\mathrm{\Delta T}{}_f=K{}_{f}m=1.86\frac{K\mathrm{kg}}{\mathrm{mol}}4\frac{\mathrm{mol}}{\mathrm{kg}}=7.44K$

This means the solution freezes at -7.44°C. However, we are cooling it to -10°C, which is 2.56°C lower than its freezing point (ΔT = 10 - 7.44 = 2.56 K). At this temperature, more ice will form, making the remaining solution more concentrated.

Step 4: Find the New Concentration at -10°C

Let 'x' grams of water separate as ice. The mass of water remaining as liquid solvent will then be (250 - x) g.

The number of moles of solute (1 mol) remains unchanged. The new molality (m') of the remaining solution must correspond to a freezing point of -10°C.

Therefore, the new depression ΔT'_f = 10 K (0°C - (-10°C) = 10 K).

We can now find the new molality required for this depression.

$\mathrm{\Delta T}{}_f\text{'}=K{}_{f}m\text{'}$
$10=1.86m\text{'}$
$m\text{'}=\frac{10}{1.86}\approx 5.376\frac{\mathrm{mol}}{\mathrm{kg}}$

Step 5: Relate New Molality to the Amount of Ice Separated

Molality is defined as moles of solute per kg of solvent. The mass of the remaining solvent is (250 - x) grams, or (250 - x)/1000 kg.

$m\text{'}=\frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$
$5.376=\frac{1}{(250-x)/1000}$
$5.376=\frac{1000}{250-x}$

Step 6: Solve for 'x'

$5.376(250-x)=1000$
$250-x=\frac{1000}{5.376}\approx 186$
$x=250-186=64$

Final Answer: 64 g of water separates as ice.

Related Concepts & Formulae

Colligative Properties: Properties of solutions that depend on the number of solute particles, not their identity. Examples include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Freezing Point Depression Formula:

$\mathrm{\Delta T}{}_f=iK{}_{f}m$

Where 'i' is the van't Hoff factor, which accounts for the dissociation of solute. For a non-dissociating solute like ethylene glycol, i = 1.

Molality (m):

$m=\frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$