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1 answer:
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Answer:
a) The work done is 10.0777 kJ
b) The water's change in internal energy is -122.1973 kJ
Explanation:
Given data:
1 mol of liquid water
T₁ = temperature = 100.9°C
P = pressure = 1 atm
Endothermic reaction
T₂ = temperature = 100°C
1 mol of water vapor
VL = volume of liquid water = 18.8 mL = 0.0188 L
VG = volume of water vapor = 30.62 L
3.25 moles of liquid water vaporizes
Q = heat added to the system = -40.7 kJ
Questions: a) Calculate the work done on or by the system, W = ?
b) Calculate the water's change in internal energy, ΔU = ?
Heat for 3.25 moles:
The work done:
The change in internal energy:
The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Using the equation,
Δ = i
m
where:
Δ = change in freezing point (unknown)
i = Van't Hoff factor
= freezing point depression constant
m = molal concentration of the solution
Molality is expressed as the number of moles of the solute per kilogram of the solvent.
Molal concentration is as follows;
MM KCl = 74.55 g/mol
molal concentration =
molal concentration = 0.1219m
Now, putting in the values to the equtaion Δ = i
m we get,
Δ = 2 × 1.86 × 0.1219
Δ = 0.4536°C
So, Δ of solution is,
Δ = 0.00°C - 0.45°C
Δ = - 0.45°C
Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Learn more about freezing point here;
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