Solution:

Given:
Conductivity of a saturated solution of CsI3 = 3.06 x 10-2 ohm m-1
Molar conductance of CsI3 = 1.53 ohm-1 m2 mol-1

To find: Ksp of CsI3

Formula:
Molar conductance = K x 1000 / C

Where,
K = Molar conductivity of the electrolyte solution at infinite dilution
C = Concentration of the solution in mol m-3

Calculation:

1. Calculation of molar conductivity at infinite dilution

At infinite dilution, the electrolyte solution dissociates completely into its constituent ions. So, the molar conductivity at infinite dilution can be calculated as the sum of the molar conductivities of the individual ions.

Molar conductivity of Cs+ ion = 82.5 ohm-1 m2 mol-1
Molar conductivity of I- ion = 76.3 ohm-1 m2 mol-1

Molar conductivity of CsI3 at infinite dilution = 3 x 82.5 + 1 x 76.3 = 323.8 ohm-1 m2 mol-1

2. Calculation of concentration of the solution

Since the solution is saturated, the concentration of CsI3 in the solution is equal to its solubility product (Ksp).

Let the solubility of CsI3 be x mol L-1
Then, the concentration of Cs+ ions and I- ions in the saturated solution will be x mol L-1 each.

Therefore, the molar concentration of CsI3 in the saturated solution = x mol L-1

3. Calculation of Ksp

Molar conductance of the saturated solution of CsI3 = 1.53 ohm-1 m2 mol-1
Concentration of CsI3 in the saturated solution = x mol L-1

Putting these values in the formula,

1.53 = K x 1000 / x

K = 1.53 x x / 1000 = 0.00153 x

But, K = [Cs+]3 [I-]

Substituting the values of [Cs+] and [I-] in terms of x,

K = (x)3 (x) = x4 mol4 L-4

Therefore, Ksp of CsI3 = x4 = (0.00153 x)4 = 3.3 x 10-13

Answer:
The Ksp of CsI3 is 3.3 x 10-13.