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

the volume of fluid displaced would equal the volume of the gas. use a gas syringe - this is a laboratory instrument that can be used to withdraw a certain volume of gas from a closed system or to measure the volume of gas from a chemical reaction.

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation: