The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (MW = 62.07 g/mol, molecularcompound) dissolved in 1.5g ofwater?

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

The temperature of the solution is -8.98 °C

Explanation:

Step 1: Data given

The normal freezing point of water is 0.00 ⁰C

Mass of ethylene glycol = 450.0 mg = 0.45 grams

Molar mass ethylene glycol = 62.07 g/mol

Mass of water = 1.5 grams = 0.0015 kg

Step 2: Calculate moles ethylene glycol

Moles ethylene glycol = mass / molar mass

Moles ethylene glycol = 0.45 grams / 62.07 g/mol

Moles ethylene glycol = 0.00725 moles

Step 3: Calculate molality

Molality = moles / mass

Molality = 0.00725 moles / 0.0015 kg

Molality = 4.83 molal

Step 4:

ΔT = i*Kf*m

⇒ i = the van't Hoff factor = 1

⇒ Kf = the freezing point depression constant = 1.86 °C/ m

⇒ m = molality = 4.83 molal

ΔT = 1 * 1.86 °C/m * 4.83 m

ΔT = 8.98

ΔT = T (pure solvent) − T (solution)

8.98 = 0.0 °C - (-8.98°C)

The temperature of the solution is -8.98 °C

2b2t

explanation:

2b2t