1.

18 g of glucose, C_(6)H_(12)O_(6) (Molar Maas=180 g mol^(1))is dissolved in 1 kg of water in a sauce pan. Al what temperature will this solution boil ? (K_(b) for water = 0.52 K kg mol^(-1), boiling point of pure water = 373.15 K )

Answer»

Solution :`Delta T_(B) = K_(b) xx m`
`T_(b) - T_(b)^(0) = 0.52 K " KG " mol^(-1)`
`xx (18 g)/(180 " g " mol^(-1) ) xx (1)/(1kg) `
`T_(b) - 372.15 = (0.52)/(10)`
`T_(b) - 373.15 = 0.052 `K
`T_(b) = 0.052 + 373.15`
`T_(b) = 373.202` K


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