Question

For an aqueous solution freezing point is – ${0.186}^{\circ }C$. Elevation of the boiling point of the same solution is ($K_f$ =1.86 $mol{e}^{-1}$ kg and $K_b$ = 0.512 $mol{e}^{-1}$ kg)

• A. ${51.2}^{\circ }C$
• B. ${5.12}^{\circ }C$
• C. ${0.512}^{\circ }C$
• D. ${0.0512}^{\circ }C$

Answer 1

The correct option is D ${0.0512}^{\circ }C$

$\frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f}\frac{\Delta T_b}{0.186}=\frac{0.512}{1.86}\Delta T_b={0.0152}^{\circ }C$