Studdy AI Logo
Math  /  Data & Statistics

QuestionEthylene glycol (C2H6O2)\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right) is a molecular compound that is used in many commercial anti-freezes. A water solution of ethylene glycol is used in vehicle radiators to lower its freezing point and thus prevent the water in the radiator from freezing. Calculate the freezing point of a solution of 552 g of ethylene glycol in 892 g of water. Kf=1.86∘C/m\mathrm{K}_{\mathrm{f}}=1.86^{\circ} \mathrm{C} / \mathrm{m}
Use the following mass numbers: C- 12 H - 1 O-16 Write the final answer in 2 decimal places.

Studdy Solution

STEP 1

1. The formula for freezing point depression is ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m, where ii is the van't Hoff factor (which is 1 for ethylene glycol, a non-electrolyte), KfK_f is the cryoscopic constant, and mm is the molality of the solution.
2. The molar mass of ethylene glycol (C2H6O2)\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right) needs to be calculated.
3. The solution consists of 552 g of ethylene glycol and 892 g of water.

STEP 2

1. Calculate the molar mass of ethylene glycol.
2. Determine the molality of the solution.
3. Calculate the freezing point depression.
4. Determine the new freezing point of the solution.

STEP 3

Calculate the molar mass of ethylene glycol (C2H6O2)\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right).
Molar mass=(2×12)+(6×1)+(2×16)=24+6+32=62 g/mol\text{Molar mass} = (2 \times 12) + (6 \times 1) + (2 \times 16) = 24 + 6 + 32 = 62 \, \text{g/mol}

STEP 4

Calculate the number of moles of ethylene glycol:
Moles of ethylene glycol=552 g62 g/mol=8.903 mol\text{Moles of ethylene glycol} = \frac{552 \, \text{g}}{62 \, \text{g/mol}} = 8.903 \, \text{mol}

STEP 5

Convert the mass of water to kilograms:
Mass of water=892 g1000=0.892 kg\text{Mass of water} = \frac{892 \, \text{g}}{1000} = 0.892 \, \text{kg}

STEP 6

Calculate the molality of the solution:
m=8.903 mol0.892 kg=9.982 mol/kgm = \frac{8.903 \, \text{mol}}{0.892 \, \text{kg}} = 9.982 \, \text{mol/kg}

STEP 7

Calculate the freezing point depression using the formula ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m:
ΔTf=1×1.86 ∘C/m×9.982 mol/kg=18.57 ∘C\Delta T_f = 1 \times 1.86 \, ^\circ \text{C/m} \times 9.982 \, \text{mol/kg} = 18.57 \, ^\circ \text{C}

STEP 8

Determine the new freezing point of the solution. The normal freezing point of water is 0∘C0^\circ \text{C}:
New freezing point=0∘C−18.57∘C=−18.57∘C\text{New freezing point} = 0^\circ \text{C} - 18.57^\circ \text{C} = -18.57^\circ \text{C}
The freezing point of the solution is:
−18.57∘C \boxed{-18.57^\circ \text{C}}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord