Which of the following aqueous solutions will | vedclass

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

Question

(A) The elevation in boiling point is given by the formula $\Delta T_b = i 	imes K_b 	imes m$. Since $K_b$ is constant for the solvent,$\Delta T_b$

Vedclass · Accepted Answer

$0.01 \ M \ Na_2SO_4$

Vedclass · Answer

(A) The elevation in boiling point is given by the formula $\Delta T_b = i 	imes K_b 	imes m$. Since $K_b$ is constant for the solvent,$\Delta T_b$ depends on the product of the van't Hoff factor $(i)$ and molality $(m)$. For $0.01 \ M \ Na_2SO_4$,$i = 3$ (as it dissociates into $2Na^+ + SO_4^{2-}$),so $i 	imes m = 3 	imes 0.01 = 0.03$. For $0.01 \ M \ KNO_3$,$i = 2$ (as it dissociates into $K^+ + NO_3^-$),so $i 	imes m = 2 	imes 0.01 = 0.02$. For $0.015 \ M$ urea,$i = 1$ (non-electrolyte),so $i 	imes m = 1 	imes 0.015 = 0.015$. For $0.015 \ M$ glucose,$i = 1$ (non-electrolyte),so $i 	imes m = 1 	imes 0.015 = 0.015$. Since $0.01 \ M \ Na_2SO_4$ has the highest value of $i 	imes m$,it will exhibit the highest boiling point.

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

Similar Questions

Calculate the molality of a solution of a nonvolatile solute having a boiling point elevation of $1.89 \ K$,if the boiling point elevation constant of the solvent is $3.15 \ K \ kg \ mol^{-1}$. (in $m$)

$A$ solution of $1$ molal concentration of a solute will have maximum boiling point elevation when the solvent is

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 $1.8 \ K \ kg \ mol^{-1}$.

If the boiling point of water is $100 \,^{\circ}C$,how many grams of $NaCl$ must be added to $500 \,g$ of water to increase the boiling point by approximately $1 \,^{\circ}C$ (in $,g$)? Given $(K_{b})_{H_2O} = 0.52 \,K \cdot kg/mol$.

The boiling point of a solution of $0.11 \ g$ of a substance in $15 \ g$ of ether was found to be $0.1 \ ^\circ C$ higher than that of the pure ether. The molecular weight of the substance will be $(K_b = 2.16 \ K \ kg \ mol^{-1})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module