Question

question 3
automotive antifreeze contains ethylene glycol, ch₂(oh)ch₂(oh), a nonvolatile nonelectrolyte, in water. calculate the freezing point of a 25.0% by mass solution of ethylene glycol in water. ( k_f = 1.86^circ c/m ) and ( k_b = 0.51^circ c/m ).
( 2.7^circ c )
( -2.7^circ c )
( -10.0^circ c )
( 10.0^circ c )

Explanation:

Step1: Define mass of solution components

Assume 100g of solution. Then mass of ethylene glycol ($m_{solute}$) is 25.0g, mass of water ($m_{solvent}$) is 75.0g = 0.075kg.

Step2: Calculate moles of ethylene glycol

Molar mass of $C_2H_6O_2$: $2(12.01) + 6(1.008) + 2(16.00) = 62.07$ g/mol.
Moles ($n$) = $\frac{25.0\mathrm{g}}{62.07\mathrm{g/mol}} \approx 0.4028$ mol.

Step3: Calculate molality (m)

Molality $m = \frac{n}{m_{solvent}(\mathrm{kg})} = \frac{0.4028\mathrm{mol}}{0.075\mathrm{kg}} \approx 5.37\mathrm{m}$. Wait, no—wait, 25% by mass means 25g solute, 75g solvent. Wait, no, wait: 25.0% by mass: mass of solute = 25.0g, solvent = 75.0g = 0.075kg. Wait, but let's recalculate moles: 25g / 62.07g/mol ≈ 0.4028 mol. Then molality m = 0.4028 mol / 0.075 kg ≈ 5.37 m? Wait, no, that can't be. Wait, no—wait, maybe I messed up. Wait, 25.0% by mass: 25g solute, 75g solvent. Wait, but freezing point depression: $\Delta T_f = iK_fm$. For nonelectrolyte, $i=1$. So $\Delta T_f = K_f \times m$. Wait, but let's check again. Wait, molar mass of ethylene glycol: C2H6O2, so 212 + 61 + 2*16 = 24 + 6 + 32 = 62 g/mol. Correct. So moles of solute: 25g / 62g/mol ≈ 0.403 mol. Mass of solvent: 75g = 0.075 kg. Molality m = 0.403 mol / 0.075 kg ≈ 5.37 m? Wait, that seems high, but maybe. Then $\Delta T_f = 1 \times 1.86 ^\circ C/m \times 5.37 m ≈ 10.0 ^\circ C$. Wait, but freezing point depression is $\Delta T_f = T_f^0 - T_f$, so $T_f = T_f^0 - \Delta T_f$. For water, $T_f^0 = 0 ^\circ C$. So $T_f = 0 - 10.0 = -10.0 ^\circ C$. Ah, there we go. So the freezing point is -10.0 °C.

Answer:

• 10.0 °C (Wait, no: the option is -10.0 °C. So the correct option is the third one: - 10.0 °C.