Question

A 0.1539 molal aqueous solution of cane sugar (molar mass = 342 g/mol) has a freezing point of 271 K, while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (molar mass = 180 g/mol) per 100 g of solution?

Answer 1

Answer:

Explanation:

Refer the below attachment.

Answer 2

Answer:

Freezing point of aqueous solution of water Tf=269.27K

Explanation:

Solution:

Given:

Molality of cane sugar solution = 0.1539

Freezing point sugar cane Tf = 271K

Freezing point of pure water T°f = 273.15K

Molality= no. of mole or solute present in one kilogram of solvent.

                                             ΔTf = Kfm

                                                 Kf=ΔTf/m=(T°f-Tf)/m

                                                 Kf=(273.15-271)K/0.1539Kg mol

                                                 Kf=13.97K Kg mol⁻¹

                                                ΔTf=Kfm

                                             T°f-Tf=13.97×(5g/180g)×(1000kg/100)

                                                  Tf=T°f-13.97×5/180×10K

                                                     =(273.15-3.88)K

Freezing point of aqueous solution of glucose Tf=269.27K

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