Question

The boiling point elevation constant for benzene is $2.57^{\circ}C/m$ The boiling point o f benzene is $81 ^{\circ}C$ Determine the boiling point of solution formed when $10 \,g$ of $C_{4}H_{12}$ is dissolved in $20\, g$ benzene

• A. $71.46$
• B. $7.14$
• C. $85.76$
• D. $88.14$

Answer 1

Answer: (D)

$\Delta T=k_{b}\cdot m$
$=k_{b}\cdot\frac{W_{B}}{M_{w_B}}\times\frac{1000}{W_{A}} \left[\begin{matrix}B \rightarrow {\text{solute}} \rightarrow C_{4} H_{12}\\ A \rightarrow {\text{solvent}} \rightarrow {\text{benzene}}\end{matrix}\right]$
$=2.57\times\frac{10}{180}\times\frac{1000}{20}$
$=7.14^{\circ}C $
Now, $\Delta T=T_{b}-T_{b}^{o}$
$\Rightarrow T_{b}=\Delta T_{b}+T_{b}^{o}$
$=7.14+81=88.14^{\circ}C$