GIVEN
Mass of solute (benzoic acid), $w_2 = 0.61$ g ; Molar mass, $M_2 = 122$ g $mol^{-1}$
Mass of solvent ($CS_2$), $w_1 = 5$ g $= 5 \times 10^{-3}$ kg
Degree of association, $\alpha = 88\% = 0.88$
$K_b = 2.3$ K kg $mol^{-1}$ ; Boiling point of pure $CS_2 = 46.2\ ^\circ C$

Step 1 - Calculate the moles of benzoic acid.
$$n = \frac{w_2}{M_2} = \frac{0.61}{122} = 0.005\ \text{mol}$$

Step 2 - Calculate the molality.
$$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.005}{5\times10^{-3}} = 1.0\ \text{mol kg}^{-1}$$

Step 3 - Calculate the van't Hoff factor for dimerisation.
Benzoic acid ASSOCIATES into dimers in $CS_2$ (a non-polar solvent) through
intermolecular hydrogen bonding:
$$2\,C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2$$
Start : 1 mol, 0
React : $\alpha$ associates, forming $\alpha/2$ mol of dimer
Total particles at equilibrium $= (1-\alpha) + \dfrac{\alpha}{2}$
$$i = \frac{(1-\alpha)+\frac{\alpha}{2}}{1} = 1 - \frac{\alpha}{2}$$
$$i = 1 - \frac{0.88}{2} = 1 - 0.44 = 0.56$$
(As expected, $i < 1$ for association.)

Step 4 - Calculate the elevation in boiling point.
$$\Delta T_b = i\,K_b\,m$$
$$\Delta T_b = 0.56 \times 2.3 \times 1.0 = 1.288\ K$$

Step 5 - Calculate the boiling point of the solution.
$$T_b = T_b^\circ + \Delta T_b = 46.2 + 1.288 = 47.488\ ^\circ C$$

ANSWER: Boiling point of the solution $\approx 47.49\ ^\circ C$ (about $47.5\ ^\circ C$)