wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

18 g of glucose, C6H12O6 is dissolved in 1 kg of water in a saucepan. At what temperature will the water boil
under atmospheric pressure? Kb for water is 0.52 K kg mol1

Open in App
Solution

Boiling point elevation of a solution is given by,
Tb=Kb×m
where,
Tb is the elevation in boiling point.
Kb is molal elevation constant.
m is molality of solution.
Molality =TbKb=Tb0.52
Given,
Mass of the glucose = 18 g
No. of moles of glucose =18180
Molality (m)=nsolute×1000Wsolventin gram

Molality =18180×10001000 mol/kg

Tb=0.52×18×1000180×1000
Boiling point of pure water, T=373 K
Tb=0.052T373 K=0.052
Thus, the temperature at which the given sugar solution boil is at 373.052 K

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon