The amount of ice that will separate out on cooling the solution containing 50g of ethylene glycol $({\mathrm{CH}}_2\mathrm{OH}{)}_2$ in 200g of water at $9.3^0\mathrm{C}$ is (${\mathrm{K}}_{\mathrm{f}}$ for water $=1.86 \mathrm{K}{ \mathrm{mol}}^{-1}$

Step 1: Calculate the depression in freezing point, ΔT_f

ΔT_f = T_0f - T_f = 0°C - (-9.3°C) = 9.3°C

Step 2: Convert the mass of water into kilograms

m_solvent = 200 g = 0.200 kg

Step 3: Calculate the number of moles of ethylene glycol

Molar mass of ethylene glycol (C₂H₆O₂):

• Carbon (C): 2 atoms × 12.01 g/mol = 24.02 g/mol
• Hydrogen (H): 6 atoms × 1.01 g/mol = 6.06 g/mol
• Oxygen (O): 2 atoms × 16.00 g/mol = 32.00 g/mol

Total molar mass: 24.02 + 6.06 + 32.00 = 62.08 g/mol

n_solute = m_solute / Molar mass = 50 g / 62.08 g/mol ≈ 0.806 mol

Step 4: Calculate the molality of the solution

m = n_solute / m_solvent = 0.806 mol / 0.200 kg = 4.03 mol/kg

Step 5: Use the freezing point depression formula

ΔT_f = K_f × m

Rearranging gives:

m = ΔT_f / K_f = 9.3°C / 1.86 K·kg·mol^-1 ≈ 5.0 mol/kg

Step 6: Calculate the mass of water that can freeze at -9.3°C

Let x be the mass of water that can freeze:

9.3 = 1.86 × (50 / 62) / (x / 1000)

Rearranging gives:

x = (1.86 × (50 / 62) × 1000) / 9.3 ≈ 161.29 g

Step 7: Calculate the amount of ice formed

Ice formed = 200 g - 161.29 g = 38.71 g