Calculate the boiling point of a solution containing 0.61 g of benzoic acid (Molar mass =122 g mol^-1 ) in 5 g of CS_2 in which it dimerises to the…

$\displaystyle \alpha = 88\% = 0.88$$\displaystyle \Delta T_b = i\,K_b\,m = i\,K_b\,\dfrac{W_B}{M_B} \times \dfrac{1000}{W_A}$$\displaystyle 0.88 = \dfrac{1-i}{1-\dfrac{1}{2}}$$\displaystyle i = 0.56$$\displaystyle T_b - T^\circ_b = 0.56 \times 2.3 \times \dfrac{0.61}{122} \times \dfrac{1000}{5}$$\displaystyle T_b - 46.2\,^\circ\mathrm{C} = 1.288\ \mathrm{K}$ or $\displaystyle ^\circ\mathrm{C}$$\displaystyle \therefore T_b = (46.2 + 1.288)\,^\circ\mathrm{C} = 47.488\,^\circ\mathrm{C}\ /\ 320.638\ \mathrm{K}$