SolutionHard
Question
One molal solution of benzoic acid in benzene boils at 81.53°C. The normal boiling point of benzene is 80.10°C. Assuming that the solute is 90% dimerized, the value of Kb for benzene is
Options
A.3.5 deg/molal
B.5.2 deg/molal
C.2.6 deg/molal
D.0.75 deg/molal
Solution
$\Delta T_{b} = K_{b}.m \Rightarrow 1.43 = K_{b} \times 1\left\lbrack 1 + 0.9\left( \frac{1}{2} - 1 \right) \right\rbrack$
$\therefore K_{b} = 2.6\text{ K/m}$
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