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?

By ankushmaster in chemistry on
No comments

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 ankushmaster

I am M.Sc (chemistry ) from punjabi university. I am a science teacher with expertise in chemistry, with 8 years of experiance in teaching. i am getting a handsome salary , now i wish to help out the people who need me UNCONDITIONALLY. and 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.

Leave a Reply

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Google photo

You are commenting using your Google account. ( Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

Cancel

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.