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

Posted byAnkush SharmaPosted inchemistry

Molality of the given solution = 0.1539m

∆Tf = 273.15 – 271 = 2.15 K

Therefore ∆Tf  = Kf .m

Kf = ∆Tf / m

Kf = 2.15/0.1539

 Now mass of the solute W2 = 5g

Molar mass of the solute M2 = 180g

Mass of the solution W1 = 100g

 Now the mass of solvent = 95g

Using the following formula:-

=4.08 K

So the freezing point of the solution will be = 273.15 – 4.08

                                                                                                = 269.07 K

Published by Ankush Sharma

I am M.Sc (chemistry ) from Punjabi University Patiala. I am a science teacher with expertise in chemistry, with 8 years of experience in teaching. Writing and blogging is my hobby, I write whenever I am free. I am constantly working on creating a new and easy way of learning the tough things in an effective way. I am constantly working to make authentic and reliable information to be shared with my students and widen the horizons of knowledge.

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