Among the following 0.1 m aqueous solutions, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The boiling point elevation is given by the formula $\Delta T_{b} = i 	imes K_{b} 	imes m$. Since the molality $(m)$ and the ebullioscopic const

Vedclass · Accepted Answer

Sodium chloride

Vedclass · Answer

(D) The boiling point elevation is given by the formula $\Delta T_{b} = i 	imes K_{b} 	imes m$. Since the molality $(m)$ and the ebullioscopic constant $(K_{b})$ are the same for all solutions,$\Delta T_{b} \propto i$,where $i$ is the van't Hoff factor.For complete ionization:$(A)$ $AlCl_3 ightarrow Al^{3+} + 3Cl^-$,so $i = 4$.$(B)$ $Al_2(SO_4)_3 ightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$.$(C)$ $K_2SO_4 ightarrow 2K^+ + SO_4^{2-}$,so $i = 3$.$(D)$ $NaCl ightarrow Na^+ + Cl^-$,so $i = 2$.The solution with the lowest $i$ value will exhibit the lowest boiling point elevation. Thus,$NaCl$ has the lowest boiling point elevation.

Among the following $0.1 \ m$ aqueous solutions,which one will exhibit the lowest boiling point elevation,assuming complete ionization of the compound in solution?

Similar Questions

Which of the following has the lowest freezing point?

If a compound $AB$ dissociates to the extent of $75\,\%$ in an aqueous solution,the molality of the solution which shows a $2.5\, K$ rise in the boiling point of the solution is $......$ molal.
(Rounded-off to the nearest integer) $[K_{b} = 0.52\, K\, kg\, mol^{-1}]$

The degree of dissociation of a $0.2 \, m$ aqueous solution of a weak acid $HX$ is $0.3$. If $K_f$ for water is $1.85 \, K \, kg \, mol^{-1}$,then the freezing point of the solution will be approximately ........... $^oC$.

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

For $0.01 \ M$ solutions of urea,$NaCl$,and $Na_2SO_4$,the ratio of their depression in freezing point is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module