Answer:
The vapor pressure of the solution is 23.636 torr
Explanation:
Where;
is the vapor pressure of the solution
is the mole fraction of the solvent
is the vapor pressure of the pure solvent
Thus,
15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl
0.67 kg of water = [(0.67*1000)/(18)]moles = 37.222 moles of H₂O
Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)
Mole fraction of solvent (water) = (37.222)/(37.222+0.261)
Mole fraction of solvent (water) = 0.993
<u>Note:</u> the vapor pressure of water at 25°C is 0.0313 atm
Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm
the vapor pressure of the solution = 0.0311 atm = 23.636 torr
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Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
