Question

The normal boiling point of toluene is ${110.7}^{o}C$ and its boiling point elevation constant is 3.32 K kg $mo{l}^{-1}$. The enthalpy of vaporization of toluene is nearly:

• A. 17 kJ $mo{l}^{-1}$
• B. 21 kJ $mo{l}^{-1}$
• C. 51 kJ $mo{l}^{-1}$
• D. 68 kJ $mo{l}^{-1}$

Answer 1

Answer: (B)

21 kJ $mo{l}^{-1}$

$K_b=\frac{R{T}_0^2}{1000L};3.32=\frac{(8.314\times {10}^{-3})\times (383.7{)}^2}{1000\times L}$
$L=0.368kJ/g$
Latent heat per mol $=0.368\times$molar mass of acetone
$=0.368\times 58=21.344kJmo{l}^{-1}$