Question

an aqueous solution freezes at -1.5°C , calculate the boiling point of the solution, kf for water=1.86k kg mole^-1.​

Answer 1

Answer:

anyone send the ans

Explanation:

an aqueous solution freezes at -1.5°C , calculate the boiling point of the solution, kf for water=1.86k kg mole^-1.

Answer 2

Answer:

Explanation:

The boiling point of an aqueous solution of a non volatile solute is 100.15°c .What is freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water and the values of Kb and Kf of water are 0.512 and 1.86 K molality-1.

Then it follows that;

For boiling point,

Tb = i x kb x M

100.15 + 273 = 1 x 0.512 x M

M = 728. 80

Then let us calculate M after adding 500 ml water

M x V = M' x 2V

M' = M/2

728.8/2

M' =364.4

Then Tf = 1 x kf x M'

         Tf = 1 x 1.86 x 364.4

          Tf = 677.8 K