Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
<h3>What is the boiling-point elevation?</h3>
Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
- Step 1: Calculate the molality of the solution.
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m
- Step 2: Calculate the boiling-point elevation.
We will use the following expression.
ΔT = Kb × m × i
ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C
where
- ΔT is the boiling-point elevation
- Kb is the ebullioscopic constant.
- b is the molality.
- i is the Van't Hoff factor (i = 2 for NaCl).
The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:
100 °C + 0.140 °C = 100.14 °C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
Learn more about boiling-point elevation here: brainly.com/question/4206205
Answer:
Q sln = 75.165 J
Explanation:
a constant pressure calorimeter:
∴ m sln = m Ba(OH)2 + m HCl
∴ molar mass Ba(OH)2 = 171.34 g/mol
∴ mol Ba(OH)2 = (0.06 L)(0.3 mol/L) = 0.018 mol
⇒ mass Ba(OH)2 = (0.018 mol)(171.34 g/mol) = 3.084 g
∴ molar mass HCl = 36.46 g/mol
∴ mol HCl = (0.06 L)(0.60 mol/L) = 0.036 mol
⇒ mass HCl = (0.036 mol)(36.46 g/mol) = 1.313 g
⇒ m sln = 3.084 g + 1.313 g = 4.3966 g
specific heat (C):
∴ C sln = C H2O = 4.18 J/g°C
∴ ΔT = 26.83°C - 22.74°C = 4.09°C
heat absorbed (Q):
⇒ Q sln = (4.3966 g)(4.18 J/g°C)(4.09°C)
⇒ Q sln = 75.165 J
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