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Scorpion4ik [409]
3 years ago
12

How much solution could be heated to boiling by the heat evolved by the dissolution of 24.0 g of naoh? (for the solution, assume

a heat capacity of 4.0 j/g⋅∘c, an initial temperature of 25.0 ∘c, a boiling point of 100.0 ∘c, and a density of 1.05 g/ml.)?
Chemistry
1 answer:
ozzi3 years ago
5 0
<span>Sodium hydroxide (NaOH) has a lattice energy of -887 kJ/mol and a heat of hydration of -932 kJ/mol.
Heat of </span><span>dissolution of 1 mol of NaOH is 887 kJ/mol - 932 kJ/mol = -45 kJ/mol.
m(NaOH) = 24 g.
n</span>(NaOH) = 24 g ÷ 40g/mol = 0,6 mol.
ΔT = 100°C - 25°C = 75°C.
Heat of dissolution of 0,6 mol of NaOH is -45 kJ/mol · 0,6 mol = -27 kJ.
27000 J = m · 4 J/g·°C · 75°C.
m(NaOH) = 90 g of solution.
d = m÷V, V = 90 g ÷ 1,05 g/ml = 85,7 ml.
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In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba ( OH ) 2 was added to 60.0 mL of 0.600 M HCl . The reaction caused the
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Answer:

Q sln = 75.165 J

Explanation:

a constant pressure calorimeter:

  • Q sln = mCΔT

∴ 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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