Answer:
0.200 m K3PO3
Explanation:
Let us remember that the freezing point depression is obtained from the formula;
ΔTf = Kf m i
Where;
Kf = freezing point constant
m = molality
i = Van't Hoff factor
The Van't Hoff factor has to do with the number of particles in solution. Let us consider the Van't Hoff factor for each specie.
0.200 m HOCH2CH2OH - 1
0.200 m Ba(NO3)2 - 3
0.200 m K3PO3 - 4
0.200 m Ca(CIO4)2 - 3
Hence, 0.200 m K3PO3 has the greatest van't Hoff factor and consequently the greatest freezing point depression.
Gay Lussac's Law

Convert:
200°C = 200 + 273 = 473 K
50°C = 50 + 273 = 323 K
Input the value:

I'd love to help, but you forgot to add the question.
It's a trigonal pyramidal, and this molecule is p<span>hosphorus trifluoride. YOU'RE WELCOME :D</span>