Question

Consider equimolal aqueous solutions of $NaHS{O}_4$ and $NaCl$ with $\Delta T_b$ and $\Delta T_b^{\prime }$ as their respective boiling point elevations. The value of $\underset{m\to 0}{\text{Lim}}\frac{\Delta T_b}{\Delta T_b^{\prime }}$ will be:

• A. $1$
• B. $1.5$
• C. $3.5$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $1.5$

Elevation in the boiling point is the colligative property and depends upon the number of particles.
As the molality approaches zero, each molecule of $NaHS{O}_4$ dissociates to give three ions and each molecule of $NaCl$ dissociates to give two ions.
$\underset{m\to 0}{Lt}\frac{\Delta T_b}{\Delta T_b^{\prime }}=\frac{3}{2}=1.5$