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Inessa [10]
3 years ago
7

The freezing point of water H2O is 0.00°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is urea. If 13.

40 grams of urea, CH4N2O (60.10 g/mol), are dissolved in 153.2 grams of water...
The molality of the solution is _____ m.
The freezing point of the solution is _____ °C.
Chemistry
1 answer:
Trava [24]3 years ago
3 0

Answer:

Molality = 1.46 molal

The freezing point of the solution = -2.72 °C

Explanation:

Step 1: Data given

The freezing point of water H2O is 0.00°C at 1 atmosphere

urea = nonelectrolyte = van't Hoff factor = 1

Mass urea = 13.40 grams

Molar mass urea = 60.1 g/mol

Mass of water = 153.2 grams

Molar mass H2O = 18.02 g/mol

Kf = 1.86 °C/m

Step 2: Calculate moles urea

Moles urea = mass urea /molar mass urea

Moles urea = 13.40 grams / 60.1 g/mol

Moles urea = 0.223 moles

Step 3: Calculate the molality

Molality = moles urea / mass water

Molality = 0.223 moles / 0.1532 kg

Molality = 1.46 molal

Step 4: Calculate the freezing point of the solution

ΔT = i * Kf * m

ΔT = 1* 1.86 °C/m * 1.46 m

ΔT = 2.72 °C

The freezing point = -2.72 °C

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Now this heat by the law of conservation of energy is equal to the heat absorbed by the water plus the heat absorbed by the calorimeter. Thus what we need to do first is to calculate those numbers.

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Note that this reaction is exothermic since heat is released which is absorbed by the water and calorimeter because ΔT  is positive.

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q H₂O = 300 g x 4.18 J/g·K x 6.1 K = 7649 J

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q cal = Ccal x  ΔT = 24.2 J/K x 6.1 K  = 1476 J

Total Heat  = q total = 7649 J + 1476 J = 7797 J

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ΔHrxn = -q total /  mol Cu(OH)₂ = 7797 J / 0.225 mol = -34,653.33 J /mol

Notice ΔHrxn is negative since it is an exothermic reaction.

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