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The boiling point of a solution made by dissolving `12.0 g` of glucose in `100 g` of water is `100.34^(@)C`. Calculate the molecular weight of glucose, `K_(b)` for water `= 0.52^(@)C//m`. |
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Answer» Using the relation for the molecular weight of a solute from elevation in boiling point, we have `Mw_(b) = K_(b)(W_(B)/(W_(A)DeltaT_(b)) xx 1000)` `= 0.52 (12/(100 xx 0.34) xx 1000)` `(DeltaT_(b) = 100.34 - 100 = 0.34^(@)C)` `rArr Mw_(B) = 183.5 g mol^(-1)` |
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