Jawaban:
a) P K₂SO₄ = 17,54 mmHg x 0,155 = 2,718 mmHg
b) Tb K₂SO₄ = 1,912°C + 100°C = 101,912°C
c) Tf K₂SO₄ = 6,621°C + 100°C = 106,621°C
Penjelasan:
Massa total = 100% = 100 gram
Massa K₂SO₄ = 64% = 64 gram
Mr K₂SO₄ = 174 g/mol
P°air = 17,54 mmHg
ρ air = 1 g/mL
Kb air = 0,52°C/kg/mol
Kf air = 1,8°C/kg/mol
a) Tekanan uap larutan
ΔP = P°air - P K₂SO₄
mol K₂SO₄ = 64 g : 174 g/mol = 0,367 mol
mol H₂O = 36 g : 18 g/mol = 2 mol
fraksi mol K₂SO₄ = 0,367 mol : (2 mol + 0,367 mol) = 0,155
P K₂SO₄ = P°air x fraksi mol K₂SO₄
P K₂SO₄ = 17,54 mmHg x 0,155 = 2,718 mmHg
b) Titik didih larutan
ΔTb = Kb x m
ΔTb = Kb x g : Mr x 1000 : p
ΔTb = 0,52°C/kg/mol x 64 g : 174 g/mol x 1000 : 100 g
ΔTb = 1,912°C
Tb K₂SO₄ = 1,912°C + 100°C = 101,912°C
c) Titik beku larutan
ΔTf = Kf x m
ΔTf = Kf x g : Mr x 1000 : p
ΔTf = 1,8°C/kg/mol x 64 g : 174 g/mol x 1000 : 100 g
ΔTf = 6,621°C
Tf K₂SO₄ = 6,621°C + 100°C = 106,621°C
[answer.2.content]
