(1) If molar mass of C₆H₅COOH is Mg mol^$$-$$1, then no. of moles of dissolved C₆H₅COOH = $${{1.22} \over M}$$ mol

$$\therefore$$ Molality (m) of the solution = $${{1.22} \over M} \times {{1000} \over {100}} = {{12.2} \over M}$$ mol kg^$$-$$1

We know, $$\Delta$$T_b = k_b $$\times$$ m

Given : $$\Delta$$T_b = 0.17 K and k_b = 1.7 K kg mol^$$-$$1

$$\therefore$$ $$0.17 = 1.7 \times {{12.2} \over M}$$ or, M = 122

$$\therefore$$ Molecular mass of C₆H₅COOH is = 122

(2) If molar mass of C₆H₅COOH is M', then no. of moles of dissolved C₆H₅COOH = $${{1.22} \over {M'}}$$ mol

Molality (m) of the solution = $${{1.22} \over {M'}} \times {{1000} \over {100}} = {{12.2} \over {M'}}$$ mol kg^$$-$$1

We know, $$\Delta$$T_b = 0.13 K and k_b = 2.6 K kg mol^$$-$$1

$$\therefore$$ $$0.13 = 2.6 \times {{12.2} \over {M'}}$$ or, M' = 244

$$\therefore$$ Molecular weight of C₆H₅COOH = 122

Now, van't Hoff factor (i) = $${{Calculated\,molecular\,weight} \over {Experimental\,molecular\,weight}} = {{122} \over {244}} = {1 \over 2}$$

$$i = {1 \over 2}$$ indicates that C₆H₅COOH undergoes complete dimerisation in benzene.