Kimia help me please trimakasiii sebelumnya ​

By di
help me please trimakasiii sebelumnya ​ 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]
Image