Calculate Van't Hoff factor for an aqueous solution of K3 [Fe(CN)6] if the degree of dissociation(α) is 0.852. What will be boiling point of this solution if its concentration is 1 molal? (Kb=0.52 Kkg/mol)

The Van't Hoff factor (i) is used to account for the effect of solute particles in solutions on colligative properties. It is defined as the ratio of moles of particles in solution to the moles of solute dissolved.

For the compound K3[Fe(CN)6], it dissociates in water as follows:

K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3-

So, theoretically, one molecule of K3[Fe(CN)6] should produce 4 particles (3 K+ ions and 1 [Fe(CN)6]3- ion).

However, since the degree of dissociation (α) is given as 0.852, not all of the K3[Fe(CN)6] molecules dissociate.

The Van't Hoff factor (i) can be calculated as follows:

i = 1 + (n-1)α

where n is the number of particles produced by the solute (in this case, 4), and α is the degree of dissociation.

i = 1 + (4-1)0.852 = 1 + 30.852 = 3.556

So, the Van't Hoff factor for this solution is 3.556.

The boiling point elevation of a solution (ΔTb) can be calculated using the formula:

ΔTb = iKbm

where i is the Van't Hoff factor, Kb is the ebullioscopic constant (in this case, 0.52 Kkg/mol), and m is the molality of the solution (in this case, 1 molal).

ΔTb = 3.5560.521 = 1.84912 K

So, the boiling point of the solution will be 1.84912 K higher than the boiling point of pure water (which is 373.15 K at standard pressure).

Therefore, the boiling point of this solution is 373.15 K + 1.84912 K = 374.99912 K.