Question

If van't Hoff factor for $KCl$ is $1.8$ . What is the boiling point of $0.75m$ solution of $KCl$ in water $K_{b}=0.5^\circ Ckg/mol$ ?

• A. $0.67^\circ C$
• B. $100.67K$
• C. $99.37^\circ C$
• D. $373.67K$

Answer 1

Answer: (D)

$ΔT_{b}=i\times K_{b}\times m$
$=1.78\times \left(\right.0.5\left.\right)\times \left(\right.0.75\left.\right)$
$ΔT_{b}=0.67^\circ C$
$T_{b}=T_{b}^\circ +ΔT_{b}$
$=100^\circ C+0.67^\circ C$
$T_{b}=100.67^\circ C$