Question 23 of 34
Q.(a)
(i) What is Van't Hoff factor 'i'?
(ii) Complete: (A) H2C=CH2 + H-Br, in the presence of Benzoyl peroxide, gives ___. (B) CH3CHO, treated with Acid dichromate, gives ___. (C) Benzene (ring structure), treated with Pt/H2, gives ___. (D) 1,4-dihydroxybenzene / hydroquinone (benzene ring with OH substituents at the 1 and 4 positions), treated with K2Cr2O7/H2SO4, gives ___.
OR
(b) Explain the purification of a solid organic compound by crystallization method.
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2020Subjective· 5mImportance★★★★★
68% · 23/34 Questions
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Start your 14-day free trial to unlock the full solution →The van't Hoff factor i corrects colligative-property calculations for dissociation or association of the solute; the four missing products are bromoethane, acetic acid, cyclohexane, and para-benzoquinone respectively.
- Van't Hoff factor, i: Colligative properties (relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure) depend on the NUMBER of solute particles present in solution, not on their identity. The simple formulas for these properties assume the solute neither dissociates nor associates. However, electrolytes dissociate into more particles (e.g. NaCl -> Na+ + Cl-, giving 2 particles per formula unit) and some solutes associate into fewer, larger particles (e.g. some carboxylic acids dimerise in non-polar solvents). To account for this, van't Hoff introduced a correction factor, i, defined as: i = (observed value of colligative property) / (calculated/normal value of colligative property, assuming no dissociation or association) Equivalently, since molar mass calculated from a colligative property is inversely related to the number of particles: i = (normal/theoretical molar mass) / (observed/experimental molar mass) For a non-electrolyte that neither dissociates nor associates, i = 1. For an electrolyte that dissociates into n ions, i approaches n (e.g. i is close to 2 for NaCl, close to 3 for CaCl2). For a solute that associates (e.g. forms dimers), i is less than 1.
- Completing the reactions: (A) H2C=CH2 + HBr, in the presence of benzoyl peroxide (a free-radical initiator): normally HBr adds to an unsymmetrical alkene by Markovnikov's rule via an ionic mechanism, but benzoyl peroxide switches the mechanism to a free-radical chain addition (the peroxide/Kharasch effect). Since ethene (H2C=CH2) is a symmetrical alkene, both mechanisms give the same single product here — bromoethane (ethyl bromide), CH3-CH2-Br. …
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