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 $10g$ of $C_4{H}_{12}$ is dissolved in $20g$ benzene

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

Answer 1

Answer: (D)

$\Delta T=k_b\cdot m$
$=k_b\cdot \frac{W_B}{M_{w_B}}\times \frac{1000}{W_A}\left[\begin{array}{c}B\to \text{solute}\to C_4{H}_{12}\\ A\to \text{solvent}\to \text{benzene}\end{array}\right]$
$=2.57\times \frac{10}{180}\times \frac{1000}{20}$
$=7.14^{\circ }C$
Now, $\Delta T=T_b-T_b^o$
$\Rightarrow T_b=\Delta T_b+T_b^o$
$=7.14+81=88.14^{\circ }C$