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