Question

The normal boiling point of toluene is ${110.7}^{0}C$ and its boiling point elevation constant is 3.32K kg $mo{1}^{-1}$. The enthalpy of vapourization of toluene is nearly:

• A. $17.0KJmo{l}^{-1}$
• B. `$34.0KJmo{l}^{-1}$
• C. $51.0KJmo{l}^{-1}$
• D. $68.0KJmo{l}^{-1}$

Answer 1

The correct option is B `$34.0KJmo{l}^{-1}$

$\Delta T_b=K_{b}m$

$m=\frac{10.7}{3.32}$

$m=3.22$

$\frac{X_A\times 1000}{(1-X_A)m_{H_{2}O}}=3.32$

$X_A=0.0548$