Which one of the following would produce maxi | vedclass

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

Question

(C) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$ and the concentration of the solute particle

Vedclass · Accepted Answer

$0.1 \ M$ barium chloride

Vedclass · Answer

(C) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$ and the concentration of the solute particles. The formula is $\Delta T_b = i 	imes K_b 	imes m$. $1$. For $0.1 \ M$ glucose (non-electrolyte),$i = 1$,so effective concentration = $0.1 	imes 1 = 0.1 \ M$. $2$. For $0.2 \ M$ sucrose (non-electrolyte),$i = 1$,so effective concentration = $0.2 	imes 1 = 0.2 \ M$. $3$. For $0.1 \ M$ barium chloride $(BaCl_2 ightarrow Ba^{2+} + 2Cl^-)$,$i = 3$,so effective concentration = $0.1 	imes 3 = 0.3 \ M$. $4$. For $0.1 \ M$ magnesium sulphate $(MgSO_4 ightarrow Mg^{2+} + SO_4^{2-})$,$i = 2$,so effective concentration = $0.1 	imes 2 = 0.2 \ M$. Since $BaCl_2$ provides the highest effective concentration of particles $(0.3 \ M)$,it will produce the maximum elevation in boiling point.

Which one of the following would produce maximum elevation in boiling point?

Similar Questions

Explain why on addition of $1 \ mol$ of $NaCl$ to $1 \ litre$ of water,the boiling point of water increases,while addition of $1 \ mol$ of methyl alcohol to one litre of water decreases its boiling point.

Which of the following aqueous solutions will exhibit the highest boiling point?

Calculate the molality of a solution containing a non-volatile solute if the boiling point elevation of the solution is $0.2 \ K$ $[K_{b} = 0.52 \ K \ kg \ mol^{-1}]$.

$58.4 \ g$ of $NaCl$ and $180 \ g$ of glucose were separately dissolved in $1000 \ mL$ of water. Identify the correct statement regarding the elevation of boiling point $(b.p.)$ of the resulting solutions.

Calculate the number of moles of nonvolatile solute dissolved in $0.3 \ kg$ solvent if $\Delta T_{b}=0.3 \ K$ and $K_{b}$ for solvent is $1.8 \ K \ kg \ mol^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module