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.
V1/T1 = V2/T2
V1= 3.5 L
T1 = 25 +273=298 K
T2 = 40.9 + 273 = 313.9 K
V2- ?
3.5 L/298 K= V2/ 313.9 K
V2 = (3.5 L*313.9 K)/298 K ≈ 3.69 L
Determine the mass of each of the following amounts. a. 1.366 mol of
NH3
, b. 0.120 mol of glucose,
C6H12O6
, c. 6.94 mol barium chloride,
BaCl2
, d. 0.005 mol of propane,
C3H8
The answer is Decomposition.