Benzophenone has a normal freezing point of +48.1 oC, with freezing point depression constant Kfpt = − 9.78 oC/m. A 0.1500 molal solution of ionizing salt had a freezing point of +44.0 oC. What is the van't Hoff (ion dissociation) constant i for this salt? (i. e., the average number of ions produced in the solution.) Report 3 significant digits.

i = 2.79

Explanation:

The excersise talks about the colligative property, freezing point depression.

Formula to calculate the freezing point of a solution is:

Freezing point of pure solvent - Freezing point of solution = m . Kf . i

Let's replace data given. (i = Van't Hoff factor, numbers of ions dissolved in solution)

48.1°C - 44°C = 0.15 m . 9.78°C/m . i

4.1°C / (0.15 m . 9.78°C/m) = i

i = 2.79

In this case, numbers of ions dissolved can decrease the freezing point of a solution, which is always lower than pure solvent.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.