0.5 m aqueous solution of a weak acid (HX) i | vedclass

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

Question

(B) The dissociation of the weak acid is represented as: $HX \rightleftharpoons H^{+} + X^{-}$.For the dissociation of an electrolyte,the van't Hoff f

Vedclass · Accepted Answer

$1.12$

Vedclass · Answer

(B) The dissociation of the weak acid is represented as: $HX ightleftharpoons H^{+} + X^{-}$.For the dissociation of an electrolyte,the van't Hoff factor $i$ is given by $i = 1 + \alpha(n-1)$,where $n$ is the number of ions produced per molecule.Here,$n = 2$ and $\alpha = 0.20$ $(20 \%)$.So,$i = 1 + 0.20(2-1) = 1.2$.The lowering in freezing point is given by $\Delta T_{f} = i 	imes K_{f} 	imes m$.Substituting the values: $\Delta T_{f} = 1.2 	imes 1.86 	imes 0.5 = 1.116 \ K \approx 1.12 \ K$.

$0.5 \ m$ aqueous solution of a weak acid $(HX)$ is $20 \%$ ionized. If $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$,the lowering in freezing point of the solution is: (in $K$)

Similar Questions

What is the expected value of $\Delta T_f$ for $1.25 \ m$ $CaCl_2$ solution if $1.25 \ m$ sucrose solution has $\Delta T_f$ value $x \ K$?

In which case is Raoult's law not applicable?

Which of the following will show maximum osmotic pressure for its $1 \, m$ solution?

$0.1 \ m$ solutions of sodium sulphate,urea,and sodium chloride are taken. The correct ratio of the elevation of boiling point of these solutions is

In an aqueous solution of potassium ferrocyanide,the degree of dissociation of the salt is $80\%$. The value of the van't Hoff factor is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module