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)
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.
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⁺
Not all HSO₄⁻ dissociates.
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) =
0.0010m SO₄²⁻
