SolutionHard
Question
How many moles of sucrose should be dissolved in 500 gms of water so as to get a solution which has a difference of 104oC between boiling point and freezing point.
(Kf = 1.86 K Kg mol-1, Kb = 0.52 K Kg mol-1)
(Kf = 1.86 K Kg mol-1, Kb = 0.52 K Kg mol-1)
Options
A.1.68
B.3.36
C.8.40
D.0.840
Solution
Boiling point of solution = boiling point + ᐃTb = 100 + ᐃTb
Freezing point of solution = freezing point - ᐃTf = 0 - ᐃTf
Difference in temperature (given) = 100 + ᐃTb - (- ᐃTf)
104 = 100 + ᐃTb + ᐃTf = 100 + molality × Kb + molality × Kf
= 100 + molality (0.52 + 1.86)
∴ Molality =
= 1.68 m
and molality =
; 1.68 = 
∴ Moles of solute =
= 0.84 moles.
Freezing point of solution = freezing point - ᐃTf = 0 - ᐃTf
Difference in temperature (given) = 100 + ᐃTb - (- ᐃTf)
104 = 100 + ᐃTb + ᐃTf = 100 + molality × Kb + molality × Kf
= 100 + molality (0.52 + 1.86)
∴ Molality =
= 1.68 mand molality =
; 1.68 = ∴ Moles of solute =
= 0.84 moles.Create a free account to view solution
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