H20. 2 of hydrogen and oxygen
Answer:The boiling point of the solution is 108° C.
Explanation:
Boiling point of pure water=T=
Boiling point of water after addition of 500 g of ethylene glycol=
Mass of water = 500g = 0.5 kg (1000 g = 1 kg)






The boiling point of the solution is 108° C.
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Experiment is performed to support or Refuge a hypothesis which is given by someone.
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Answer:
0.0010m SO₄²⁻
Explanation:
The freezing point depression due the addition of a solute into a pure solvent follows the equation:
ΔT = Kf×m×i (1)
<em>Where ΔT are °C that freezing point decreases (273.15K - 272.47K = 0.68K = 0.68°C). Kf is the constant of freezing point depression (1.86°C/m), m is molality of the solution (0.1778m) and i is Van't Hoff factor.</em>
Van't Hoff factor could be understood as in how many one mole of the solute (sulfuric acid, H₂SO₄), is dissociated.
H₂SO₄ dissociates as follows:
H₂SO₄ → HSO₄⁻ + H⁺
HSO₄⁻ ⇄ SO₄²⁻ + H⁺
<em>Not all HSO₄⁻ dissociates.</em>
1 Mole of H₂SO₄ dissociates in 1 mole of H⁺+ 1 mole of HSO₄⁻ + X moles of SO₄²⁻= 2 + X
Replacing in (1):
0.68°C = 1.86°C/m×0.1778m×i
2.056 = i
Moles of SO₄²⁻ are 2.056 - 2 = 0.056moles SO₄²⁻.
If 1 mole has a concentration of 0.1778m, 0.056moles are:
0.056moles ₓ (0.1778m / 1mole) =
<h3>0.0010m SO₄²⁻</h3>