The molal elevation constant of water is 0.52 | vedclass

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

Question

(B) For $KCl$,the van't Hoff factor $i = 2$ because it dissociates as $KCl \rightarrow K^+ + Cl^-$.The formula for elevation in boiling point is $\Del

Vedclass · Accepted Answer

$101.04$

Vedclass · Answer

(B) For $KCl$,the van't Hoff factor $i = 2$ because it dissociates as $KCl ightarrow K^+ + Cl^-$.The formula for elevation in boiling point is $\Delta T_b = i 	imes K_b 	imes m$.Given $i = 2$,$K_b = 0.52 \, ^{\circ}C \, kg \, mol^{-1}$,and $m = 1.0 \, m$.$\Delta T_b = 2 	imes 0.52 	imes 1.0 = 1.04 \, ^{\circ}C$.The boiling point of the solution $T_b = T_b^{\circ} + \Delta T_b$.Since the boiling point of pure water $T_b^{\circ} = 100 \, ^{\circ}C$,we have $T_b = 100 + 1.04 = 101.04 \, ^{\circ}C$.

The molal elevation constant of water is $0.52 \, ^{\circ}C \, kg \, mol^{-1}$. The boiling point of $1.0 \, m$ aqueous $KCl$ solution (assuming complete dissociation of $KCl$) should be $......... \, ^{\circ}C$.

Similar Questions

Calculate the amount of $KCl$ $(M_w = 74.5 \ g \ mol^{-1})$ in grams required to depress the freezing point of $1000 \ g$ of water by $2 \ K$. (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$) (in $.0$)

Which of the following compounds has a Van't Hoff factor $i$ equal to $2$ for a dilute solution?

In which of the following solutions are ions present?

For a dilute solution of $K_3[Fe(CN)_6]$,the van't Hoff factor $(i)$ is ..... [assuming $100\%$ ionization]

At $25 \ ^oC$,the highest osmotic pressure is exhibited by $0.1 \ M$ solution of

Vedclass Test Series

Exam Paper Generator

Online Exam Module