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

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

Answer 2

Answer:

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

Explanation:

$\Delta T_f=K_f\times m$

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

$\Delta T_f=1.86^oC\times 0.50 m$

$\Delta T_f=0.93^oC$

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.