sebanyak 36 gram glukosa (C6H1206 Mr = 180 g/mol) di larutkan 250 gram air . jika diketahui Kb air =0,52C kg/mol, tentukan titik didih larutan
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Pertanyaan
sebanyak 36 gram glukosa (C6H1206 Mr = 180 g/mol) di larutkan 250 gram air . jika diketahui Kb air =0,52C kg/mol, tentukan titik didih larutan
1 Jawaban
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1. Jawaban ionkovalen
delta Tb = m Kb
= 36/180 x 1000/250 x 0,52
= 0,416 C
Tb larutan = 100,416°C