An aq solution freezes point _0.2°C Whate is the molality of the solution determine 1 , elevation in boiling point 2 , lowering of vapour pressure at 298 k given that kf= 1.86°C kb = 0.512°C and vapour pressure of water at 298 k is 23.756
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Explanation:
The freezing point of pure water is 0 deg C and that of the aqueous solution is 0.186
0
C.
The depression in the freezing point ΔT
f
=0.186−0=0.186
o
C
ΔT
f
=K
f
×m
0.186
o
C=1.86
o
Cm×m
m=0.1 m
The aqueous solution is 1 molal.
The elevation in the boiling point
ΔT
b
=K
b
×m
ΔT
b
=0.512
o
Cm×0.1 m
ΔT
b
=0.0512
o
C
The boiling point of pure water is 100 deg C. The boiling point of the aqueous solution is 100+0.512=100.0512
o
C
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