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Question
The molecular weight of an organic compound is 58 g mol-1. What will be the boiling point of a solution containing 48 grams of the solute in 1200 grams of water?
[Kb for water = 0.513°C kg mole-1; Boiling point of water = 100°C]
Solution
`"M"_"B" = 58` g mol-1
`"T"_"b" = ?`
`"T"_"b"^0 = 100^circ"C"`
`"W"_"B" = 48 "g"`
`"W"_"A" = 1200 "g"`
`"K"_"b" = 0.513^circ"C" "kg mol"^-1`
We know that
`Δ"T"_"b" = "K"_"b" xx ("W"_"B" xx 1000)/("W"_"A" xx "M"_"B")`
= `(0.513 xx 48 xx 1000)/(1200 xx 58)`
= 0.353°C
`"T"_"b" = "T"_"b"^0 + Δ"T"_"b" = 100 + 0.353 = 100.353^circ"C"`
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