Diketahui glukosa(mr=180) dan natrium sulfat(mr=142)masing masing 2 di larutkan dalam 100 gram air (kb=0,5 dan kf=1,86)tentukan titik didih dan titik beku larutan​

Jawaban:

natrium sulfat

Tb larutan = 104,35°C

Tf larutan = -16,182°C

glukosa

Tb larutan = 102,75°C

Tf larutan = -0,23°C

Penjelasan:

Diketahui:

misalkan

massa glukosa = 100 gram

massa natrium sulfat = 100 gram

p = 100 gram

Kb = 0,5°C/m

Kf = 1,86°C/m

Ditanyakan:

Tb larutan

Tf larutan

Jawab:

Natrium sulfat = Na₂SO₄

Na₂SO₄ —> 2Na⁺ + SO₄²⁻

n = 3

ΔTb = m/Mr x 1000/p x Kb x i

ΔTb = 100/342 x 1000/100 x 0,5 x i

ΔTb = 0,29 x 10 x 0,5 x 3

ΔTb = 4,35°C

Tb larutan = 100 + ΔTb

Tb larutan = 100 + 4,35

Tb larutan = 104,35°C

ΔTf = m/Mr x 1000/p x Kb x i

ΔTf = 100/342 x 1000/100 x 1,86 x i

ΔTf = 0,29 x 10 x 1,86 x 3

ΔTf = 16,182°C

Tf larutan = 0 – ΔTf

Tf larutan = 0 – 16,182

Tf larutan = -16,182°C

Glukosa

ΔTb = m/Mr x 1000/p x Kb

ΔTb = 100/180 x 1000/100 x 0,5

ΔTb = 0,55 x 10 x 0,5

ΔTb = 2,75°C

Tb larutan = 100 + ΔTb

Tb larutan = 100 + 2,75

Tb larutan = 102,75°C

ΔTf = m/Mr x 1000/p x Kb

ΔTf = 100/180 x 1000/100 x 1,86

ΔTf = 0,55 x 10 x 1,86

ΔTf = 10,23°C

Tf larutan = 0 – ΔTf

Tf larutan = 0 – 10,23

Tf larutan = -0,23°C

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