Question

18 gram glucose is dissolved in 100 gram of water calculate the freezing point of the solution​

Answer 1

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

Answer 2

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

                                            $=\frac{18/180}{100/1000}$  

                                           = 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.