experiement 5 - 0.60 0.65 0.70 2.0 2.5 3.0 3.5 4.0 4.5 5.0...

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naphthalene Solvent, 10g = 0.01 kg O benzophenon Solute, Molar mass = 182.2 gmol -1
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0 2 4 6 8 10 78 80 82 84 86 88 90 92 Temperature ( o C) Time (min) Temperature against Time ( Pure Naphthalene) Freezing Point : 80 o C
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0 2 4 6 8 10 72 74 76 78 80 82 84 86 88 90 Temperature ( o C) Time (min) Temperature against Time ( Naphthalene + 0.6 benzophenon) Freezing Point : 78 o C so, T f0.6 = 80 – 78 = 2 o C
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0 2 4 6 8 10 70 75 80 85 90 95 Temperature ( o C) Time (min ) Temperature against Time ( Naphthalene + 1.2 benzophenon) Freezing Point : 75 o C so, T f1.2 = 80 – 75 = 5 o C
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T f = K f m m = molality e naphthalen of kg n benzopheno of solvent of kg solute mol mol = = m 33 . 0 kg 0.01 10 3.3 molality moles 10 3 . 3 2 . 182 6 . 0 n benzopheno of g 0.6 of moles 3 - 3 1 × = × = = - - gmol g For 1.2 g of benzophenon, Molarity = 0.66 m
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0.30 0.35 0.40 0.45 0.50 0.55
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Unformatted text preview: 0.60 0.65 0.70 2.0 2.5 3.0 3.5 4.0 4.5 5.0 f ( o C) molality (m) lot T f against molality and calculate K f , K f is the slope of the gra Let says your Kf = 9 o C kgmol-1 2 4 6 8 10 70 75 80 85 90 95 Temperature ( o C) Time (min) Temperature against Time ( Naphthalene + 0.6 unknown) Freezing Point : 76.5 o C so, T f unknown = 80 76.5 = 3.5 o C 1 3 3-1-o 154 10 89 . 3 6 . moles mass Mr 10 3.89 solute of moles 0.01 solute of moles 389 . solvent of kg solute of moles m 389 . 9 5 . 3 Ckgmol 9 5 . 3--= = = = = = = = = = gmol m m m K T f f...
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This note was uploaded on 06/21/2009 for the course CHEM 130 taught by Professor Kia during the Spring '09 term at Abraham Baldwin Agricultural College.

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experiement 5 - 0.60 0.65 0.70 2.0 2.5 3.0 3.5 4.0 4.5 5.0...

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