Question

For a 0.50 molal solution of sucrose in water, calculate the freezing point and the boiling point of the solution.

Answer 1

Answer:

-0.93 °C; 100.26 °C  

Step-by-step explanation:

(a) Freezing point depression

The formula for the freezing point depression ΔT_f is

ΔT_f = iKf·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For sucrose,

    Sucrose (s)   ⟶   sucrose (aq)

1 mole sucrose ⟶ 1 mol particles     i = 1

ΔT_f = 1 × 1.86 × 0.50

ΔT_f = 0.93 °C

 T_f = T_f° - ΔT_f  

 T_f = 0.00 – 0.93  

 T_f = -0.93 °C

(b) Boiling point elevation

The formula for the boiling point elevation ΔTb is

ΔTb = iKb·b

ΔTb = 1 × 0.512 × 0.50

ΔTb = 0.256 °C

 Tb = Tb° + ΔTb  

 Tb = 100.00 + 0.256  

 Tb = 100.26 °C