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nikdorinn [45]
3 years ago
7

A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing point depression of

the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
7 0
From the equation; ΔTf = Kf × m
Where, Kf for water = 1.853 K kg/mole; m is the molarity = number of solute/amount of solvent in kg.
Glucose is the solute whose molecular mass is 180 g/mole and water is the solvent. 
Moles of solute = 15.5/180 = 0.0861 moles
Amount of solvent in kg = 245/1000 = 0.245 Kg
Therefore; molarity = 0.0861/0.245 = 0.3515 moles/Kg
Therefore; ΔTf = 1.853 × 0.3515 = 0.6513 K
Hence; the depression in freezing point is 0.6513 
The freezing point of solution will therefore be;
= 273 - 0.6513 = 272.3487 K

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Answer:

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Explanation:

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3 years ago
A 72.0-gram piece of metal at 96.0 °C is placed in 130.0 g of water in a calorimeter at 25.5 °C. The final temperature in the ca
tekilochka [14]

Answer: The answer is S = 0.1528 cal/g °C

Explanation:

By the law of conservation of energy, energy is neither created nor destroyed.

So, energy lost by metal pieces is equal to the energy gained by water in the calorimeter.

Specific heat of water is 1 cal/g °C

⇒ heat energy  Q = mSΔT, where m = mass of a substance

                                                        S = specific heat

                                                        ΔT = change in temperature

Now, the heat lost by metal piece, Q = 72×S×(96-31)

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Heat gained by water, Q = 130×1×(31-25.5)

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