Question

What will be the boiling point of $30 \,g$ benzene containing $0.75\, g$ of benzoic acid which undergoes $75\%$ dimerization? Boiling point of pure benzene $= 80.1^{\circ}C$ and $K_b = 2.53 \,K mol^{-1}\, kg$

• A. $90.52^{\circ}C$
• B. $104.35^{\circ}C$
• C. $76.12 ^{\circ}C$
• D. $80.42 ^{\circ}C$

Answer 1

Answer: (D)

$80.42 ^{\circ}C$

Answer 2

No. of moles of benzene $=\frac{30}{78}$
No. of moles of benzoic acid $=\frac{0.75}{122}=6.15\times10^{-3}$
$\Delta\,T_{b}=i K_{b}\,m$
$i=1-\alpha+\frac{\alpha}{n}$
$=1-0.75+\frac{0.75}{2}$
$=0.625$
$\Delta T_{b}=0.625\times2.53\times\frac{6.15\times10^{-3}}{30}\times1000$
$\Delta T_{b}=0.324$
$\Delta T_{b}=T_{b}-T_{b}^{\circ}$
$0.324=T_{b}-80.1\,^{\circ}C$
$T_{b}=80.42\,^{\circ}C$