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 increases

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.