Question

The unit of ebullioscopic constant is _______________.

• A. $molkg{K}^{-1}or{K}^{-1}\left(molality\right)$
• B. $Kmolk{g}^{-1}orK\left(molality\right)$
• C. $Kkgmo{l}^{-1}orK(molality{)}^{-1}$
  
• D. $kgmo{l}^{-1}{K}^{-1}or{K}^{-1}(molality{)}^{-1}$
  

Answer 1

The correct option is C $Kkgmo{l}^{-1}orK(molality{)}^{-1}$


Ebullioscopic constant relates molality to the boiling point elevation. This relation is:

$\Delta T_b=i.K_b.m$

$K_b=\frac{\Delta T_b}{i.m}$

Where, $\Delta T_b$ = elevation in the boiling point

i= van’t Hoff factor, m is the molality

$K_b$= Ebullioscopic constant

Unit of $K_b$

$K_b=\frac{K}{mol/kg}$

$K_b=Kkgmo{l}^{-1}$

As the unit of molality is $molk{g}^{-1}$
Thus, the unit of an ebullioscopic constant is $Kkgmo{l}^{-1}$ 𝑜𝑟 $K(molality{)}^{-1}$.

Hence, option (A) is correct.