The freezing point of equimolal aqueous solution will be highest for:

Question

The freezing point of equimolal aqueous solution will be highest for: (A) C6H5N+H3Cl- (aniline hydrochloride) (B) Ca ( NO 3)2 (C) La ( NO 3)3 (D) C6H12O6 (Glucose)

Tardigrade Admin · Accepted Answer

Δ Tf propto m× i m = molality, i = Van't Hoff factor Except glucose, all other electrolytes are undergoing dissociation so (.i>1.) . Glucose is non electrolyte, i=1 . So, it has least value of Δ Tf hence, its freezing point is maximum.

Q. The freezing point of equimolal aqueous solution will be highest for:

$C_{6} H_{5} N^{+} H_{3} C l^{-}$ (aniline hydrochloride)

$C a (N O_{3})_{2}$

$L a (N O_{3})_{3}$

$C_{6} H_{12} O_{6}$ (Glucose)

$Δ T_{f} \propto m \times i$
m = molality, i = Van't Hoff factor
Except glucose, all other electrolytes are undergoing dissociation so $(i > 1)$ . Glucose is non electrolyte, $i = 1$ . So, it has least value of $Δ T_{f}$ hence, its freezing point is maximum.

Q. The freezing point of equimolal aqueous solution will be highest for:

C6​H5​N+H3​Cl− (aniline hydrochloride)

Ca(NO3​)2​

La(NO3​)3​

C6​H12​O6​ (Glucose)

ΔTf​∝m×i m = molality, i = Van't Hoff factor Except glucose, all other electrolytes are undergoing dissociation so (i>1) . Glucose is non electrolyte, i=1 . So, it has least value of ΔTf​ hence, its freezing point is maximum.

$C_{6} H_{5} N^{+} H_{3} C l^{-}$ (aniline hydrochloride)

$C a (N O_{3})_{2}$

$L a (N O_{3})_{3}$

$C_{6} H_{12} O_{6}$ (Glucose)

$Δ T_{f} \propto m \times i$
m = molality, i = Van't Hoff factor
Except glucose, all other electrolytes are undergoing dissociation so $(i > 1)$ . Glucose is non electrolyte, $i = 1$ . So, it has least value of $Δ T_{f}$ hence, its freezing point is maximum.