Question

# 18g of glucose (C6H12O6) is dissolved in 1 kg of water in a saucepan. At what temperature will the water boil (at 1 atm)? Kb for water is 0.52Kkgmol−1.

A

373.202K

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B

365.202K

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C

370.202K

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D

473.202K

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Solution

## The correct option is A 373.202K The given values are: Wsolute=18g Wsolvent=1kg kb=0.52kkgmol−1 First we calculate elevation in the boiling point of solution. ΔTb=Kb×WsoluteMwsolute×Wsolvent 0.52×18180×1=0.052K Since water boils at 373.15 K at 1 atm pressure, therefore the boiling point of solution will be TB=Tb+ΔTb=373.15+0.052=373.202K Thus, the boiling point of solution is 373.202 K .

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