If you use 3.68mol of sucrose (C12H22O12) and dissolve this into 2.50kg of water, what will be the change in the freezing point of your solution? Assume the Kf of water is -1.86°C/m

-2.74°C

Explanation:

From the formula;

∆T= Kf m i

Where Kf= freezing constant of water

m= molality of the solution = number of moles of solute/ mass of solvent in kilogram

Since number of moles of solute = 3.68 mol and mass of solvent= 2.5Kg

m= 3.68/2.5 = 1.472

Then the freezing constant Kf= -1.86 as given in the question

Since the solution is not ionic, the Vant Hoft factor (i) = 0

Hence;

∆T= Kf m

∆T= -1.86 × 1.472

∆T= -2.74°C

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it