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Brut [27]
3 years ago
5

An antifreeze solution is prepared containing 40.0 g of ethylene glycol (molar mass = 62.0 g/mol) in 60.0 g of water. Calculate

the freezing point of this solution.
A. 120.1⁰C
B. 0.518 ⁰C
C. 20.1⁰C
D. -20.1⁰C
Chemistry
1 answer:
suter [353]3 years ago
6 0
Kf = 1.86

Number of moles :

Molar mass <span> ethylene glycol = 62.0 g/mol

40.0 / 62.0 => 0.645 moles 

Molality :

mass of solvent = 60.0 / 1000 => 0.06 kg

m = moles / kg solvent

m = </span><span>0.645 / 0.06

= 10.75 m

</span>Δt = Kf x molality

<span>Δt = 1.86 x 10.75 => </span>20.1ºC
<span>
Answer C

hope this helps!</span>
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Answer: The maximum mass of sucrose you can add is 4158.95 grams.

Explanation:

This is an example of freezing point depression. The formula for calculating this is the following:

ΔT = Kf . b . i

ΔT is the temperature depression

Kf is the cryoscopic constant that is unique for each solvent

b is the molality of the solution (moles of solute per kg of solvent)

i is the Vant Hoff factor

The freezing point of water is 0°C. ΔT equals inicial temperature - final temperature, so it's 0°- (-15°)= 273K - 258K = 15K

The Kf for water is known to be 1.853 K. Kg /mol

i is the number of particles the molecule is split to when ionized. Because sucrose doesn't ionize, its Vant Hoff factor is 1.

If we clear b from the ecuation:

b = ΔT/ Kf . i

b = 15K/ 1.853 K. Kg/mol . 1

b= 8.1 mol/kg

If we can add 8.1 moles to a kg of water before it freezes, we use cross multiplication to calculate how many we can add to 1.5 kg. The answer is 12.15.

The weight of a mole of sucrose is 342.3 grams. So the weight of 12.15 moles of sucrose is 4158.95g.

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             = 1.51 × 10²³ particles

                           

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