18 gram glucose is dissolved in 100 gram of water calculate the freezing point of the solution
Answers
Answer:
∆Tf=kf×molality here ∆Tf= freezing point depression (Tsolvent-Tsolution) and kf is molal freezing point depression constant and it's value for water is 1.86°c/m and molality = no.of mole of solute ×1000/weight of solvent ,,, ∆Tf = 1.86×18×1000/180×100 after solving it ∆Tf is 1.86°c and then ∆Tf =T solvent- T solution we know that T solvent means water freezing point is 0°c then 1.86=0-Tsolution ,,, T solution= -1.86°c ans
Given: mass of solute(glucose) = 18 gm and mass of solvent(water) = 100 gm
To Find: Freezing point of solution
Formula used: ΔTf = i × Kf × m
where, ΔTf = lowering in freezing point
i = dissociation/association constant
Kf = freezing point depression constant
m = molality
To calculate molality-
Formula - moles of solute/mass of solvent in kg
= 1 m
ΔTf = i × Kf × m
= 1 × 1.86 × 1
( i = 1 for non-ionic solutions and Kf value is usually mentioned in the question)
ΔTf = 1.86
ΔTf i.e. depression in freezing point = Freezing point of water - freezing point of the solution
1.86 = 0 - F
F = 0 - 1.86
F = - 1.86°C
Therefore, by adding glucose in water the freezing point of the solution becomes - 1.86°C.