On dissolving 6g glucose in 50g water, 0.34℃ elevation in boiling point is recorded. Calculate molal elevation constant of water.
First right answer will be marked as brainliest and wrong or spam answer are reported and deleted.
Answers
Boiling point of a solution is found to be 100.34
o
C. Boiling point of pure
water is 100
o
C.
The elevation in the boiling point ΔT
b
=100.34−100=0.34
o
C.
12 gm glucose (molecular weight 180 g/mol) is dissolved in 100 gm water.
The number of moles of glucose =
180g/mol
12g
=0.0667mol
Mass of water =100g×
1000g
1kg
=0.100kg
Molality of solution m=
0.100kg
0.0667mol
=0.667mol/kg
The elevation in the boiling point ΔT
b
=K
b
×m
0.34
o
C=K
b
×0.667mol/kg
K
b
=0.51
o
Ckg/mol
The molal elevation constant for water is 0.51
o
Ckg/mol.
Answer~
Given:-
- Glucose = 6g
- Water = 50g
- Elevation in boiling point = 0.34 °C
To find:-
- Molal elevation constant =?
_______________________________
Volume of solvent = 50g = 0.05kg
No of moles of solute = Given mass / Molar mass
No of moles = 6g /180g = 0.033 moles
First find the molality of the solvent
- Molality = no of mole of solute/ Volume of solvent (kg)
- Molality = 0.033/0.05
- Molality = 0.66 moles/kg
Formula used:-
∆Tf = kf x m
Value of m is 0.66 moles /kg and ∆Tf is 0.34°C and we have to find the value of kf.
=> ∆Tf = kf x m
=> 0.34°C = kf x 0.66
=> kf = 0.35/0.66
=> kf = 0.5151 = 51.51 x 10^-2