Question

There is 50% dimer formation of benzoic acid $\left(C_6{H}_{5}COOH\right)$ in benzene solution.
$2C_6{H}_{5}COOH\rightleftharpoons (C_6{H}_{5}COOH{)}_2$,
Hence the abnormal molecular weight of benzoic acid ( theoretical value = 122 g $mo{l}^{-1}$) is:

• A. 61 g $mo{l}^{-1}$
• B. 244 g $mo{l}^{-1}$
• C. 163 g $mo{l}^{-1}$
• D. 81 g $mo{l}^{-1}$

Answer 1

Answer: (B)

244 g $mo{l}^{-1}$

$2C_6{H}_{5}COOH\rightleftharpoons (C_6{H}_{5}COOH{)}_2$

initial $1$ $-$

At $e{q}^m$ $1-\alpha$ $\frac{\alpha }{n}$

$\alpha$ =degree of dissociation of reactant

Total moles=$1-\alpha +\frac{\alpha }{n}$

$\alpha =\frac{i-1}{\frac{1}{n}-1}$

$i=1-\alpha +\frac{\alpha }{n}$

$i$=Vantoff factor

$\frac{\alpha }{n}=0.500$ (given)

$\alpha =0.50\times 2$ ($n$=2)

$\alpha =1$

$i=X-X+\frac{1}{2}=\frac{1}{2}$

$\frac{1}{2}=\frac{Normalmolarmass}{abnormalmolarmass}$

$2\times Normalmolarmass=abnormalmolarmass$

$2\times 122=244$