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jasenka [17]
3 years ago
5

Assuming ideal behavior, what is the freezing point of a 0.030 m solution of Al2(SO4)3 ?

Chemistry
1 answer:
Andrews [41]3 years ago
3 0
The freezing point of ideal solutions lower respect the freezing point of the pure solvent following the freezing depression formula:

ΔT = i * Kf * m

Where:

-  i is the vant'Hoff, which is the number of partilces (ions in this case)  that are the solute forms when dissociate.

- kf is cryoscopic constant of the pure solvent (water because we assume it is an aqueous solution).

Then kf = - 1.86 °C / mol

- and m is the molality, m = 0.030 m

i = 2 + 3 = 5, given the one molecule of Al2 (SO4)3 produces 2 ions of Al(+) and 3 ions of SO4 (2-).

Then, ΔT = 5 * 1.86°C/m * 0.030 m = 0.279 °C.

That is the depression of the freezing point which you have to subtract from the freezing point of the pure water => 0°C - 0.279°C = - 0.279 °C.

Answer: - 0.279°C 

 
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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
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Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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