What is the freezing point of a 0.19 m solution of glucose, c6h12o6, in water? (kf for water is 1.858°c/m.) a. –0.18°c b. 0.18°c c. 0.35°c d. –0.35°c e. –0.71°c?

- d. -0.35 degree C is the right answer.

Solution:- depression in freezing point, delta Tf formula is...

delta Tf = i x m x kf

where, i is Van't hoff factor. glucose is covalent compound and does not dissociate to give ions and so the value of i for this is 1.

m is molaity and kf is the molal freezing point constant.

Let's plug in the given values in the formula..

delta Tf = 1 x 0.19 x 1.858

delta Tf = 0.35

Pure water freezes at 0 degree C.

So, freezing point of solution = 0 - 0.35 = -0.35 degree C