Question

One molal solution of benzoic acid in benzene boils at $81.53{}^{\circ }C$. The normal boiling point of benzene is $80.10{}^{\circ }C.$ Assuming that the solute is $90\%$ dimerized, the value of $K_b$ i.e. ebullioscopic constant for benzene is:

• A. $3.5{}^{\circ }C/molal$
• B. $5.2{}^{\circ }C/molal$
• C. $2.6{}^{\circ }C/molal$
• D. $0.75{}^{\circ }C/molal$

Answer 1

The correct option is C $2.6{}^{\circ }C/molal$

$\text{Elevated boiling point}={81.53}^{\circ }C=T_b$
$\text{Normal boiling point}={80.10}^{\circ }C=T_b^o$
$\Delta T_b=T_b-T_b^0=(81.53-80.10{)}^{\circ }C={1.43}^{\circ }C$
Benzoic acid associates and forms dimer.
$n=2$
$\alpha =\frac{1-i}{1-\frac{1}{n}}$
Here i is Vant Hoff Factor
$0.9=\frac{1-i}{1-\frac{1}{2}}$
$i=0.55$
$\therefore$ For elevation in boiling point,
$\Delta T_b=i\times k_b\times m$

$\Rightarrow K_b=\frac{1.43}{1\times 0.55}={2.6}^{\circ }C/molal$