Question

if 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2)

Answer 1

This can be solved with the equation for freezing point depression:
[delta]T_f = K_fsolvent * m_solute

[delta]T_f is change in freezing point (will be a negative change).
K_fsolvent is the cryoscopic constant of the solvent; for water it is 1.86.
m_solute is the molality of the solute, LiF; molality is mols of solute/kg of solvent

So. you have (52.5g LiF)/(26 g/mol) = 2.02 mol LiF
you have .306 kg (KG!!!!) of water. 2.02 / .306 = 6.69.

BUT 6.69 IS THE WRONG MOLALITY. LiF ionizes to Li+ and F- in water, so there are TWICE as many mols of solute, or (2.02*2)= 4.04 total mols of solute.
So now, 4.04/.306 = 13.20 molal LiF.

It's easy from here: 13.20*1.86 (as per the equation above) = 24.55degrees celsius.
standard Freezing point of water = 0 degrees Celsius
0 - 24.55= -24.55