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1.22 g of benzoic acid is dissolved in 100 g of acetone and 100
g of benzene separately. Boiling point of the solution in acetone
increases by 0.17ºC, while that of, in the benzene increases by 0.13ºC;
Kb for acetone and benzene is 1.7 K kg mol-1 and 2.6 K kg mol-1. Find
molecular weight of benzoic acid in two cases and justify your answer.
1.22 g csatksbd vEy dks i`Fkd:i ls 100 g,lhVksu rFkk 100 g csathu esa
tcfd csathu esa 0.13ºC rd DoFkukad c<+rk gSA,lhVksu rFkk csathu ds fy,
Kb Øe'k 1.7 K kg mol-1 rFkk 2.6 K kg mol-1 gSA nksuksa csatkWbd vEy dk vkf.od Hkkj Kkr djrs gq,] vius mÙkj

Show answer and solution
(i) In first case, çFke voLFkk esa]
ΔTb=K0×m=K0×Wt ˙of solute  Mol. wt. of solute ×1000 wt. of solvent \Delta T_{b} = K_{0} \times m = K_{0} \times \frac{W_{t}\.\ of\ solute\ }{\ Mol.\ wt.\ of\ solute\ } \times \frac{1000}{\ wt.\ of\ solvent\ }
or 0.17=1.7×1.22M×100×10−30.17 = 1.7 \times \frac{1.22}{M \times 100 \times 10^{- 3}}
or M = 122 g/mole
Thus the benzoic acid exists as a monomer in acetone
vr csUtkWbd vEy],lhVksu esa,dyd ds:i aA
(ii) In second case, f}rh; voLFkk esa]
ΔTb=K0×W t.of solute  Mol.wt.of solute ×1000 wt. of solvent \Delta T_{b} = K_{0} \times \frac{W\ t.of\ solute\ }{\ Mol.wt.of\ solute\ } \times \frac{1000}{\ wt.\ of\ solvent\ }
or 0.13=2.6×1.22M×100×10−30.13 = 2.6 \times \frac{1.22}{M \times 100 \times 10^{- 3}}
or M` = 224
Double molecular weight of benzoic acid (244) in acetone solution
indicates that benzoic acid exists as a dimer in acetone.
lhVksu csUtkWbd vEy dk nqxquk v.kqHkkj (244) fd csUtksbd vEy,lhVksu

Source: Grass-G / CH-27 Chemistry Paper 2 15 Jan.docx

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