Question

Questions number 5

5. An aqueous solution of glucose (C6H1206) freezes at -5.25°C (Kf for water = 1.86 °C/n, Kb for water = 0.512 °C/n, freezing point of water = 0.00°C; boiling point of water = 100.0°C) a) Calculate the molality of the solution (S-points) Showu wrand worke b) Calculate the boiling point of the solution. (5 points) Show all set up and work !

Answer 1

Given

F.P of water = 0.00 ⁰ C

F.P of glucose solution = - 5.25 ⁰ C

K _f = 1.86  ⁰ C /m

K _b =0.512  ⁰ C /m

B.P of solvent = 100.0 ⁰ C

a)

We have relation, T _f = K _f m

Where, T _f is a depression in freezing point = F.P of solvent - F. P of solution, K _f is a freezing point constant of solvent & m is a molality of solution.

$\therefore$ F.P of solvent - F. P of solution = K _f m

Substituting given values in above relation, we get 0.00 ⁰ C - ( - 5.25 ⁰ C ) = 1.86 ⁰ C / m molality of solution

5.25 ⁰ C = 1.86 ⁰ C / m molality of solution

molality of Glucose solution = 5.25 ⁰ C / 1.86 ⁰ C / m

molality of Glucose solution = 2.822 mol / kg

ANSWER : molality of Glucose solution = 2.82 m

b)

We have relation, B.P of solution - B.P of solvent = K _b m

Where , K _b is boiling point constant of solvent and m is molality of solution.

Substituting given values in above relation, we get

B.P of solution - ( 100.0 ⁰ C ) = 0.512 ⁰ C / m 2.822 m

B.P of solution - ( 100.0 ⁰ C ) = 1.445

B.P of solution = 100.0 ⁰ C + 1.445 ⁰ C

B.P of solution = 101.4 ⁰ C

ANSWER : B.P of Glucose solution = 101.4 ⁰ C