SOLUTIONS 1.53 TABLE 9. MOLAL ELEVATION CONSTANTS OF SOME SOLVENTS Solvent B.P. (K) Molal elevation constant (Kb) (K kg mol–1 or K molarity–1) Water 373 Ethyl alcohol 351·5 0·52 Benzene 353·3 1·20 Chloroform 334·4 2·53 Carbon tetrachloride 350·0 3·63 Diethyl ether 307·8 5·03 Cyclohexane 353·74 2·02 Acetic acid 391·1 2·79 2·93 Determination of Observed Molar Mass of Solute We know that, DTb = Kb × m m (Molality) 1 WB / MB 1 WB 2 1000 WA / 1000 MB 2 WA \\ DTb 1 Kb 2 m 1 Kb 2 WB 2 1000 MB 2 WA MB 1 Kb 2 WB 2 1000 . 3Tb 2 WA In the above relation mass of solvent (WA) is in grams. In case, it is taken in kilograms (kg), then MB = Kb 1 WB 2Tb 1 WA All other values being known the observed molecular mass of the solute can be calculated. It can match the normal molecular mass provided (i) the solute is non–volatile (ii) the solution is dilute and obeys Raoult's Law. (iii) the solute behaves normally in solution i.e., it does not undergo any association or dissociation leading to the change in number of particles. 122345627589 7 35723 4 DTb = Kb × m DTb 1 Kb 2 WB ; Kb 1 MRTb22 MB 2 WA 3Hvap Example 63 The boiling point of a solution containing 50 gm of a non-volatile solute in 1 kg of solvent is 0·5° higher than that of the pure solvent. Determine the molecular mass of the solute (Given molecular mass of solvent = 78 g mol–1 and Kb for solvent = 2·53 km–1) (Haryana Board 2003) Solution. MB 1 Kb 2 WB 3Tb 2 WA WB = 50 g ; WA = 1 kg ; DTb = 0·5 K ; Kb = 2·53 km–1 or 2·53 K kg mol–1 MB 1 (2.53 K kg mol31 ) 2 (50 g) = 253 g mol–1 ( 0.5 K ) 2 (1 kg) Example 64 On dissolving 3·24 g of sulphur in 40 g of benzene, boiling point of the solution was higher than that of benzene by 0·81K. The molal elevation constant (Kb) for benzene is 2·53 K kg mol–1. What is the molecular formula of sulphur (Atomic mass of S = 32 g mol–1) ?
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