Question

What is the expected freezing point of a 0.50 m solution of na2so4 in water kf for water is 1.86°c/m?

Answer 1

The expected freezing point of a 0.50 m solution of sodium sulfate is -0.93°C.

Step-by-step explanation:

`\Delta T_f=K_f* m`

where,

`\Delta T_f` =depression in freezing point =

`K_f` = freezing point constant

m = molality

we have :

`K_f` =1.86°C/m ,

m = 0.50 m

`\Delta T_f=1.86^oC* 0.50 m`

`\Delta T_f=0.93^oC`

Freezing point of pure water = T = 0°C

Freezing point of solution =
`T_f`

`\Delta T_f=T-T_f`

`T_f=T-\Delta T_f=0^oC-0.93^oC=-0.93^oC`

The expected freezing point of a 0.50 m solution of sodium sulfate is -0.93°C.

Answer 2

Colligative properties calculations are used for this type of problem. Calculations are as follows:


ΔT(freezing point) = (Kf)m
ΔT(freezing point) = 1.86 °C kg / mol (0.50 mol/kg)
ΔT(freezing point) = 0.93 °C
Tf - T = 0.93 °C
T = -0.93 °C