Question

_________ has the highest boiling point.

• A. 0.2 m $Ca{Cl}_2$
• B. 0.2 m $NaCl$
• C. 0.1 m $Al{Cl}_3$
• D. 0.2 m ${CH}_{3}OH$
• E. 0.2 m $Na{C}_2{H}_3{O}_2$

Answer 1

The correct option is A 0.2 m $Ca{Cl}_2$

$\Delta T_b=i\times k_{b}m$
$i=\frac{Dissociated-moles}{Initial-moles}$
For (A) : 0.2 m $CaC{l}_2$, $i=\frac{3\times 0.2}{0.2}=\frac{0.6}{0.2}=3$
Now, as the molarity of all is same before dissociation, hence, $k_{b}m$ value for all is same, as $i$ value for $CaC{l}_2$ is highest. Hence, $i\times K_{b}m$ value for $CaC{l}_2$ is highest , hence as per formula, the elevation in boiling point is highest for the $CaC{l}_2$ case.

Here $AlC{l}_3$ has i=4 but its molar concentration is low.
Hence, Highest boiling point is for $0.2m$ $CaC{l}_2$.