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massa urea = 6 gr
massa air = 200 gr
P = 1 atm
Kf = 1,86 °C/m
Mr urea = 60
Jawab:
ΔTf = m . kf
ΔTf = (massa urea . 1000 . Kf) / (Mr urea . massa air)
ΔTf = (6 . 1000 . 1,86 ) / (60 . 200)
ΔTf = 0,93 ° C
ΔTf = Tf° - Tf
Tf = Tf° - ΔTf
Tf = 0 - 0,93
Tf = - 0,93 ° C
Jika yang diminta penurunan titik beku, maka jawabannya adalah ΔTf
Jika yang diminta titik beku larutan, maka jawabannya adalah Tf.