Question

12.2 gm of benzoic acid (m=122) in 100gm $H_{2}O$ has an elevation of a boiling point of 0.27. $k_b=0.540$kg/mol. If there is 100% polymerization, the number of molecules of a benzoic acid associated state is:

Answer 1

The number of moles of benzoic acid are
$\frac{\text{mass}}{\text{molar mass}}=\frac{12.2}{122}=0.1$ moles.
The molality of benzoic acid is
$\frac{\text{number of moles of benzoic acid}}{\text{mass of water in kg}}=\frac{0.1}{\frac{100}{1000}}=1$ m
The expression for the elevation of boiling point is
$\Delta T_b=i\times K_b\times m$
Here, i is the van't Hoff's factor.
$i=\frac{\Delta T_b}{K_b\times m}$
$i=\frac{0.27}{0.54\times 1}=0.5$
Since there is 100% polymerization, the degree of polymerization is 1 (or 100%).
$\therefore \alpha =1$
The number of benzoic acid molecules that combine to give one associated molecule is
$n=\frac{\alpha }{\alpha -1+i}$
$n=\frac{1}{1-1+0.5}$
$n=2$