0.5 molal aqueous solution of a weak acid (HX | vedclass

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

Question

(B) The depression in freezing point is given by the formula: $\Delta T_{f} = i 	imes K_{f} 	imes m$. For the dissociation of weak acid $HX ightle

Vedclass · Accepted Answer

$1.12$

Vedclass · Answer

(B) The depression in freezing point is given by the formula: $\Delta T_{f} = i 	imes K_{f} 	imes m$. For the dissociation of weak acid $HX ightleftharpoons H^{+} + X^{-}$,the van't Hoff factor $i$ is calculated as $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation. Given $\alpha = 20\% = 0.2$,so $i = 1 + 0.2 = 1.2$. Given molality $m = 0.5\, mol\, kg^{-1}$ and $K_{f} = 1.86\, K\, kg\, mol^{-1}$. Substituting these values: $\Delta T_{f} = 1.2 	imes 1.86 	imes 0.5 = 1.116\, K \approx 1.12\, K$.

$0.5$ molal aqueous solution of a weak acid $(HX)$ is $20\%$ ionised. If $K_f$ for water is $1.86\, K\, kg\, mol^{-1}$,the lowering in freezing point of the solution is $........ K$.

Similar Questions

$17.4\% \text{ (mass/volume) } K_2SO_4$ solution at $27^\circ C$ is isotonic to $5.85\% \text{ (mass/volume) } NaCl$ solution at $27^\circ C$. If $NaCl$ is $100\%$ ionized,what is the $\%$ ionization of $K_2SO_4$ in aqueous solution? [Atomic weights: $K = 39, Na = 23, S = 32, O = 16, Cl = 35.5$]

What is the freezing point (in $^\circ C$) of a solution containing $0.1 \ g$ of $K_3[Fe(CN)_6]$ (molar mass = $329 \ g \ mol^{-1}$) in $100 \ g$ of water? (Given $K_f = 1.86 \ K \ kg \ mol^{-1}$)

Which of the following aqueous solutions has the highest boiling point?

Which of the following will have the highest boiling point at $1 \ atm$ pressure?

Of the following $0.10 \ m$ aqueous solutions,which one will exhibit the largest freezing point depression?

Vedclass Test Series

Exam Paper Generator

Online Exam Module