The unit of ebullioscopic constant is .

Question

The unit of ebullioscopic constant is . (A) K kg mol-1 or K ( textmolality)-1 (B) mol kg K-1 or K-1 ( textmolality) (C)  kg mol-1 K-1 or K-1 ( textmolality)-1 (D) K mol kg-1 or K text(molality)

Tardigrade Admin · Accepted Answer

As we know from elevation in boiling point that U Δ Tb = Kb m Unit of Kb = ( textUnit of Δ Tb/ textUnit of m) = (K/mol kg-1) = K mol-1 kg

Q. The unit of ebullioscopic constant is ______.

$K k g m o l^{- 1}$ or $K (molality)^{- 1}$

$m o l k g K^{- 1}$ or $K^{- 1} (molality)$

$k g m o l^{- 1} K^{- 1}$ or $K^{- 1} (molality)^{- 1}$

$K m o l k g^{- 1}$ or $K (molality)$

As we know from elevation in boiling point that $U$

$Δ T_{b} = K_{b} m$

Unit of $K_{b} = \frac{Unit of Δ T _{b}}{Unit of m}$

$= \frac{K}{m o l k g ^{- 1}} = K m o l^{- 1} k g$

Q. The unit of ebullioscopic constant is ______.

Kkgmol−1 or K(molality)−1

molkgK−1 or K−1(molality)

kgmol−1K−1 or K−1(molality)−1

Kmolkg−1 or K(molality)

As we know from elevation in boiling point that U ΔTb​=Kb​m Unit of Kb​=Unit ofmUnit of ΔTb​​ =molkg−1K​=Kmol−1kg

$K k g m o l^{- 1}$ or $K (molality)^{- 1}$

$m o l k g K^{- 1}$ or $K^{- 1} (molality)$

$k g m o l^{- 1} K^{- 1}$ or $K^{- 1} (molality)^{- 1}$

$K m o l k g^{- 1}$ or $K (molality)$

As we know from elevation in boiling point that $U$

$Δ T_{b} = K_{b} m$

Unit of $K_{b} = \frac{Unit of Δ T _{b}}{Unit of m}$

$= \frac{K}{m o l k g ^{- 1}} = K m o l^{- 1} k g$