Question

What is the expected freezing point of a 0.50 m solution of li2so4 in water? Kf for water is 1.86 ∘c/m.

Answer 1

Answer: -2.79°C

Explanation:

Supposing complete ionization:

Li2SO4 → 2 Li{+} + SO4{2-}  [three ions]

(0.50 m Li2SO4) x 3 = 1.50 m ions  

(1.50 m) x (1.86 °C/m) = 2.79°C change

0°C - 2.79°C = -2.79°C

Answer 2

Answer:

-2.79°C is the expected freezing point of a 0.50 m solution of lithium sulfate.

Explanation:

Freezing point of water = T =0°C

Melting point of sample = $T_f$

Depression in freezing point = $\Delta T_f$

Depression in freezing point  is also given by formula:

$\Delta T_f=i\times K_f\times m$

$K_f$ = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor  

We have: $K_f$ = 40°C kg/mol

m = 0.50 m

i = 3 (lithium sulfate is an ionic compound)

$\Delta T_f=3\times 1.86^oC kg/mol\times 0.50 m=2.79^oC$

$\Delta T_f=T- T_f$

$T_f=T-\Delta T_f=0^oC-2.79^oC=-2.79^oC$

-2.79°C is the expected freezing point of a 0.50 m solution of lithium sulfate.