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Anton [14]
3 years ago
15

sebanyak 36 gram glukosa (C6H12O6,M=180 g/mol) dilarutkan dalam 250 gram air.jika diketahui Kb air= 0,52° kg/mol.tentukan titik

didih larutan tersebut
Chemistry
1 answer:
likoan [24]3 years ago
4 0
Gr terlarut = 36 gr 
<span>Mr terlarut = 180 </span>
<span>gr pelarut = 250 gr </span>
<span>Kb air = 0,52 °C kg/mol </span>

<span>Tb larutan = ........? </span>
<span>--------------------------------------... </span>
<span>ΔTb = Kb.m.i </span>
<span>ΔTb = Kb. (gr t / Mr t) . (1000/ gr p) .i </span>
<span>ΔTb = 0,52 x (36/180) x (1000/250) x 1 </span>
<span>ΔTb = 0,416 °C </span>

<span>Tb = 100 + ΔTb </span>
<span>Tb = 100 + 0,416 </span>
<span>Tb = 100,416 °C</span>
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